Chapter 13 Exponents and Powers

Given

The number 256.

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To Find

Express 256 as a power of 2, i.e., find $x$ such that $2^x = 256$.

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Solution

We can find the required power by performing prime factorization of 256 using the base 2.

We repeatedly divide 256 by 2:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

From the factorization, we can see that 256 is the product of 8 factors of 2.

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Therefore, 256 can be expressed as a power of 2 as:

$256 = 2^8$

The required expression is $2^8$.

Given

The two numbers in exponential form are $2^3$ and $3^2$.

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To Find

Which of the two numbers, $2^3$ or $3^2$, is greater.

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Solution

To compare the two numbers, we need to calculate their values.

First, calculate the value of $2^3$:

$2^3 = 2 \times 2 \times 2 = 4 \times 2 = 8$

Next, calculate the value of $3^2$:

$3^2 = 3 \times 3 = 9$

Now, compare the calculated values, 8 and 9.

We know that $9 > 8$.

Since $9 = 3^2$ and $8 = 2^3$, we can conclude that:

$3^2 > 2^3$

Therefore, $3^2$ is greater than $2^3$.

Given

The two numbers in exponential form are $8^2$ and $2^8$.

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To Find

Which of the two numbers, $8^2$ or $2^8$, is greater.

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Solution

To compare the two numbers, we need to calculate their values.

First, calculate the value of $8^2$:

$8^2 = 8 \times 8 = 64$

Next, calculate the value of $2^8$:

$2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 4 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 8 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 16 \times 2 \times 2 \times 2 \times 2$

$= 32 \times 2 \times 2 \times 2$

$= 64 \times 2 \times 2$

$= 128 \times 2$

$= 256$

Now, compare the calculated values, 64 and 256.

We know that $256 > 64$.

Since $256 = 2^8$ and $64 = 8^2$, we can conclude that:

$2^8 > 8^2$

Therefore, $2^8$ is greater than $8^2$.

Given

The expressions are $a^3b^2$, $a^2b^3$, $b^2a^3$, and $b^3a^2$.

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To Find

1. Expand each expression.

2. Determine if all the expanded forms are the same.

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Solution

Let's expand each expression:

(i) $a^3b^2$

Expansion: $a^3b^2 = a \times a \times a \times b \times b$

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(ii) $a^2b^3$

Expansion: $a^2b^3 = a \times a \times b \times b \times b$

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(iii) $b^2a^3$

Expansion: $b^2a^3 = b \times b \times a \times a \times a$

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(iv) $b^3a^2$

Expansion: $b^3a^2 = b \times b \times b \times a \times a$

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Now, let's compare the expanded forms.

Comparing $a^3b^2$ and $b^2a^3$:

$a^3b^2 = a \times a \times a \times b \times b$

$b^2a^3 = b \times b \times a \times a \times a$

Since multiplication is commutative (the order of factors does not change the product), rearranging the factors in $b^2a^3$ gives $a \times a \times a \times b \times b$.

Thus, $a^3b^2 = b^2a^3$.

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Comparing $a^2b^3$ and $b^3a^2$:

$a^2b^3 = a \times a \times b \times b \times b$

$b^3a^2 = b \times b \times b \times a \times a$

Similarly, due to the commutative property of multiplication, rearranging the factors in $b^3a^2$ gives $a \times a \times b \times b \times b$.

Thus, $a^2b^3 = b^3a^2$.

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Comparing $a^3b^2$ and $a^2b^3$:

$a^3b^2 = a \times a \times a \times b \times b$ (Three factors of $a$, two factors of $b$)

$a^2b^3 = a \times a \times b \times b \times b$ (Two factors of $a$, three factors of $b$)

These expressions have different numbers of factors for $a$ and $b$. Therefore, $a^3b^2 \neq a^2b^3$ (unless $a=b$ or one of them is 0 or 1, but generally they are different).

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Conclusion:

We found that $a^3b^2 = b^2a^3$ and $a^2b^3 = b^3a^2$.

However, $a^3b^2$ is not the same as $a^2b^3$.

Therefore, not all the given expressions are the same.

Given

The numbers 72, 432, 1000, and 16000.

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To Find

Express each number as a product of powers of its prime factors.

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Solution

We use the method of prime factorization for each number.

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(i) 72

Performing prime factorization for 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 72 is $2 \times 2 \times 2 \times 3 \times 3$.

Writing this in terms of powers of prime factors:

$72 = 2^3 \times 3^2$

Therefore, 72 expressed as a product of powers of prime factors is $2^3 \times 3^2$.

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(ii) 432

Performing prime factorization for 432:

$\begin{array}{c|cc} 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 432 is $2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

Writing this in terms of powers of prime factors:

$432 = 2^4 \times 3^3$

Therefore, 432 expressed as a product of powers of prime factors is $2^4 \times 3^3$.

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(iii) 1000

Performing prime factorization for 1000:

$\begin{array}{c|cc} 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 1000 is $2 \times 2 \times 2 \times 5 \times 5 \times 5$.

Writing this in terms of powers of prime factors:

$1000 = 2^3 \times 5^3$

Therefore, 1000 expressed as a product of powers of prime factors is $2^3 \times 5^3$.

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(iv) 16000

Performing prime factorization for 16000:

$\begin{array}{c|cc} 2 & 16000 \\ \hline 2 & 8000 \\ \hline 2 & 4000 \\ \hline 2 & 2000 \\ \hline 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 16000 is $2 \times 2 \times 2 \times 2 \times 2 \times \ $$ 2 \times 2 \times \ $$ 5 \times 5 \ $$ \times 5$.

Writing this in terms of powers of prime factors:

$16000 = 2^7 \times 5^3$

Therefore, 16000 expressed as a product of powers of prime factors is $2^7 \times 5^3$.

We need to evaluate each expression involving powers.

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(i) $(1)^5$

Solution:

$(1)^5 = 1 \times 1 \times 1 \times 1 \times 1$

Multiplying 1 by itself any number of times results in 1.

$(1)^5 = 1$

Therefore, $(1)^5 = \mathbf{1}$.

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(ii) $(–1)^3$

Solution:

$(-1)^3 = (-1) \times (-1) \times (-1)$

First, multiply the first two factors: $(-1) \times (-1) = 1$.

$= (1) \times (-1)$

Now, multiply the result by the last factor: $1 \times (-1) = -1$.

Therefore, $(–1)^3 = \mathbf{-1}$. (Note: -1 raised to any odd power is -1).

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(iii) $(–1)^4$

Solution:

$(-1)^4 = (-1) \times (-1) \times (-1) \times (-1)$

Group the factors: $[(-1) \times (-1)] \times [(-1) \times (-1)]$

$= (1) \times (1)$

$= 1$

Therefore, $(–1)^4 = \mathbf{1}$. (Note: -1 raised to any even power is 1).

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(iv) $(–10)^3$

Solution:

$(-10)^3 = (-10) \times (-10) \times (-10)$

First, multiply the first two factors: $(-10) \times (-10) = 100$.

$= (100) \times (-10)$

Now, multiply the result by the last factor: $100 \times (-10) = -1000$.

Therefore, $(–10)^3 = \mathbf{-1000}$.

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(v) $(–5)^4$

Solution:

$(-5)^4 = (-5) \times (-5) \times (-5) \times (-5)$

Group the factors: $[(-5) \times (-5)] \times [(-5) \times (-5)]$

$= (25) \times (25)$

$= 625$

Therefore, $(–5)^4 = \mathbf{625}$.

(i) $2^6$

Solution:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 4 \times 2 \times 2 \times 2 \times 2$

$= 8 \times 2 \times 2 \times 2$

$= 16 \times 2 \times 2$

$= 32 \times 2$

$= 64$

Therefore, the value of $2^6$ is 64.

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(ii) $9^3$

Solution:

$9^3 = 9 \times 9 \times 9$

$= 81 \times 9$

$= 729$

Therefore, the value of $9^3$ is 729.

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(iii) $11^2$

Solution:

$11^2 = 11 \times 11$

$= 121$

Therefore, the value of $11^2$ is 121.

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(iv) $5^4$

Solution:

$5^4 = 5 \times 5 \times 5 \times 5$

$= 25 \times 5 \times 5$

$= 125 \times 5$

$= 625$

Therefore, the value of $5^4$ is 625.

Exponential form represents repeated multiplication of the same base. The base is the number being multiplied, and the exponent indicates how many times the base is multiplied by itself.

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(i) $6 × 6 × 6 × 6$

Solution:

The base is 6.

The base 6 is multiplied by itself 4 times.

Therefore, the exponential form is $6^4$.

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(ii) $t × t$

Solution:

The base is $t$.

The base $t$ is multiplied by itself 2 times.

Therefore, the exponential form is $t^2$.

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(iii) $b × b × b × b$

Solution:

The base is $b$.

The base $b$ is multiplied by itself 4 times.

Therefore, the exponential form is $b^4$.

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(iv) $5 × 5 × 7 × 7 × 7$

Solution:

The base 5 is multiplied by itself 2 times, which is $5^2$.

The base 7 is multiplied by itself 3 times, which is $7^3$.

Therefore, the exponential form is $5^2 \times 7^3$.

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(v) $2 × 2 × a × a$

Solution:

The base 2 is multiplied by itself 2 times, which is $2^2$.

The base $a$ is multiplied by itself 2 times, which is $a^2$.

Therefore, the exponential form is $2^2 \times a^2$.

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(vi) $a × a × a × c × c × c × c × d$

Solution:

The base $a$ is multiplied by itself 3 times, which is $a^3$.

The base $c$ is multiplied by itself 4 times, which is $c^4$.

The base $d$ appears 1 time, which is $d^1$ or simply $d$.

Therefore, the exponential form is $a^3 \times c^4 \times d$.

To express a number using exponential notation, we find its prime factorization and write it in terms of powers of its prime factors.

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(i) 512

Solution:

We perform prime factorization of 512.

$\begin{array}{c|cc} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

The prime factorization is $512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Since the factor 2 is repeated 9 times, the exponential notation is:

$512 = \mathbf{2^9}$

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(ii) 343

Solution:

We perform prime factorization of 343.

$\begin{array}{c|cc} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorization is $343 = 7 \times 7 \times 7$.

Since the factor 7 is repeated 3 times, the exponential notation is:

$343 = \mathbf{7^3}$

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(iii) 729

Solution:

We perform prime factorization of 729.

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization is $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

Since the factor 3 is repeated 6 times, the exponential notation is:

$729 = \mathbf{3^6}$

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(iv) 3125

Solution:

We perform prime factorization of 3125.

$\begin{array}{c|cc} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization is $3125 = 5 \times 5 \times 5 \times 5 \times 5$.

Since the factor 5 is repeated 5 times, the exponential notation is:

$3125 = \mathbf{5^5}$

(i) $4^3$ or $3^4$

Solution:

Calculate the value of $4^3$: $4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$.

Calculate the value of $3^4$: $3^4 = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81$.

Comparing the values: $81 > 64$.

Therefore, $3^4$ is greater than $4^3$.

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(ii) $5^3$ or $3^5$

Solution:

Calculate the value of $5^3$: $5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$.

Calculate the value of $3^5$: $3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 \ $$ = 81 \times 3 = 243$.

Comparing the values: $243 > 125$.

Therefore, $3^5$ is greater than $5^3$.

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(iii) $2^8$ or $8^2$

Solution:

Calculate the value of $2^8$: $2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$.

Calculate the value of $8^2$: $8^2 = 8 \times 8 = 64$.

Comparing the values: $256 > 64$.

Therefore, $2^8$ is greater than $8^2$.

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(iv) $100^2$ or $2^{100}$

Solution:

Calculate the value of $100^2$: $100^2 = 100 \times 100 = 10,000$.

Estimate the value of $2^{100}$:

$2^{10} = 1024$.

$2^{100} = (2^{10})^{10} = (1024)^{10}$.

Clearly, $(1024)^{10}$ is much larger than $100^2 = 10,000$.

Alternatively,

$100^2 = (10^2)^2 = 10^4$.

$2^{100} = (2^{10})^{10} = (1024)^{10} \approx (10^3)^{10} = 10^{30}$.

Since $10^{30} > 10^4$, we can conclude $2^{100} > 100^2$.

Therefore, $2^{100}$ is greater than $100^2$.

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(v) $2^{10}$ or $10^2$

Solution:

Calculate the value of $2^{10}$: $2^{10} = 1024$.

Calculate the value of $10^2$: $10^2 = 10 \times 10 = 100$.

Comparing the values: $1024 > 100$.

Therefore, $2^{10}$ is greater than $10^2$.

We will use prime factorization to express each number as a product of powers of its prime factors.

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(i) 648

Solution:

Performing prime factorization for 648:

$\begin{array}{c|cc} 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 648 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$.

Writing this in terms of powers of prime factors:

$648 = 2^3 \times 3^4$

Therefore, 648 expressed as a product of powers of prime factors is $2^3 \times 3^4$.

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(ii) 405

Solution:

Performing prime factorization for 405:

$\begin{array}{c|cc} 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 405 is $3 \times 3 \times 3 \times 3 \times 5$.

Writing this in terms of powers of prime factors:

$405 = 3^4 \times 5^1$ or $3^4 \times 5$.

Therefore, 405 expressed as a product of powers of prime factors is $3^4 \times 5$.

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(iii) 540

Solution:

Performing prime factorization for 540:

$\begin{array}{c|cc} 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 540 is $2 \times 2 \times 3 \times 3 \times 3 \times 5$.

Writing this in terms of powers of prime factors:

$540 = 2^2 \times 3^3 \times 5^1$ or $2^2 \times 3^3 \times 5$.

Therefore, 540 expressed as a product of powers of prime factors is $2^2 \times 3^3 \times 5$.

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(iv) 3,600

Solution:

Performing prime factorization for 3,600:

$\begin{array}{c|cc} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 3,600 is $2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$.

Writing this in terms of powers of prime factors:

$3600 = 2^4 \times 3^2 \times 5^2$

Therefore, 3,600 expressed as a product of powers of prime factors is $2^4 \times 3^2 \times 5^2$.

(i) $2 \times 10^3$

Solution:

First, calculate $10^3$: $10^3 = 10 \times 10 \times 10 = 1000$.

Now, multiply by 2: $2 \times 1000 = 2000$.

Therefore, $2 \times 10^3 = \mathbf{2000}$.

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(ii) $7^2 \times 2^2$

Solution:

Calculate $7^2$: $7^2 = 7 \times 7 = 49$.

Calculate $2^2$: $2^2 = 2 \times 2 = 4$.

Now, multiply the results: $49 \times 4 = 196$.

Therefore, $7^2 \times 2^2 = \mathbf{196}$.

(Alternatively, using laws of exponents: $7^2 \times 2^2 = (7 \times 2)^2 = 14^2 = 196$.)

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(iii) $2^3 \times 5$

Solution:

Calculate $2^3$: $2^3 = 2 \times 2 \times 2 = 8$.

Now, multiply by 5: $8 \times 5 = 40$.

Therefore, $2^3 \times 5 = \mathbf{40}$.

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(iv) $3 \times 4^4$

Solution:

Calculate $4^4$: $4^4 = 4 \times 4 \times 4 \times 4 = 16 \times 16 = 256$.

Now, multiply by 3: $3 \times 256 = 768$.

Therefore, $3 \times 4^4 = \mathbf{768}$.

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(v) $0 \times 10^2$

Solution:

Any number multiplied by 0 is 0.

Calculate $10^2$: $10^2 = 100$.

$0 \times 100 = 0$.

Therefore, $0 \times 10^2 = \mathbf{0}$.

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(vi) $5^2 \times 3^3$

Solution:

Calculate $5^2$: $5^2 = 5 \times 5 = 25$.

Calculate $3^3$: $3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$.

Now, multiply the results: $25 \times 27$.

$\begin{array}{cc}& & 2 & 7 \\ \times & & 2 & 5 \\ \hline & 1 & 3 & 5 \\ & 5 & 4 & \times \\ \hline & 6 & 7 & 5 \\ \hline \end{array}$

$25 \times 27 = 675$.

Therefore, $5^2 \times 3^3 = \mathbf{675}$.

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(vii) $2^4 \times 3^2$

Solution:

Calculate $2^4$: $2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Calculate $3^2$: $3^2 = 3 \times 3 = 9$.

Now, multiply the results: $16 \times 9 = 144$.

Therefore, $2^4 \times 3^2 = \mathbf{144}$.

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(viii) $3^2 \times 10^4$

Solution:

Calculate $3^2$: $3^2 = 3 \times 3 = 9$.

Calculate $10^4$: $10^4 = 10 \times 10 \times 10 \times 10 = 10000$.

Now, multiply the results: $9 \times 10000 = 90000$.

Therefore, $3^2 \times 10^4 = \mathbf{90000}$.

(i) $(– 4)^3$

Solution:

$(-4)^3 = (-4) \times (-4) \times (-4)$

$= (16) \times (-4)$

$= -64$

Therefore, $(-4)^3 = \mathbf{-64}$.

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(ii) $(–3) × (–2)^3$

Solution:

First, calculate $(-2)^3$:

$(-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$.

Now, multiply by $(-3)$:

$(-3) \times (-8)$

$= 24$

Therefore, $(-3) \times (-2)^3 = \mathbf{24}$.

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(iii) $(–3)^2 × (–5)^2$

Solution:

Calculate $(-3)^2$: $(-3)^2 = (-3) \times (-3) = 9$.

Calculate $(-5)^2$: $(-5)^2 = (-5) \times (-5) = 25$.

Now, multiply the results:

$9 \times 25 = 225$.

Therefore, $(-3)^2 \times (-5)^2 = \mathbf{225}$.

Alternate Solution:

Using the law of exponents $a^m \times b^m = (a \times b)^m$:

$(-3)^2 \times (-5)^2 = [(-3) \times (-5)]^2$

$= [15]^2$

$= 15 \times 15 = 225$.

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(iv) $(–2)^3 × (–10)^3$

Solution:

Calculate $(-2)^3$: $(-2)^3 = (-2) \times (-2) \times (-2) = -8$.

Calculate $(-10)^3$: $(-10)^3 = (-10) \times (-10) \times (-10) \ $$ = 100 \times (-10) \ $$ = -1000$.

Now, multiply the results:

$(-8) \times (-1000) = 8000$.

Therefore, $(-2)^3 \times (-10)^3 = \mathbf{8000}$.

Alternate Solution:

Using the law of exponents $a^m \times b^m = (a \times b)^m$:

$(-2)^3 \times (-10)^3 = [(-2) \times (-10)]^3$

$= [20]^3$

$= 20 \times 20 \times 20 = 400 \times 20 = 8000$.

(i) Compare $2.7 \times 10^{12}$ and $1.5 \times 10^8$

Given

The two numbers are $2.7 \times 10^{12}$ and $1.5 \times 10^8$.

To Find

Which number is greater.

Solution

To compare numbers in scientific notation ($a \times 10^n$), we first compare the exponents ($n$).

The exponent for the first number is $12$.

The exponent for the second number is $8$.

Since $12 > 8$, the number with the exponent 12 will be greater than the number with the exponent 8.

$10^{12}$ is significantly larger than $10^8$.

Therefore, $2.7 \times 10^{12} > 1.5 \times 10^8$.

$2.7 \times 10^{12}$ is greater than $1.5 \times 10^8$.

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(ii) Compare $4 \times 10^{14}$ and $3 \times 10^{17}$

Given

The two numbers are $4 \times 10^{14}$ and $3 \times 10^{17}$.

To Find

Which number is greater.

Solution

We first compare the exponents of 10.

The exponent for the first number is $14$.

The exponent for the second number is $17$.

Since $17 > 14$, the number with the exponent 17 will be greater than the number with the exponent 14.

$10^{17}$ is significantly larger than $10^{14}$.

Therefore, $3 \times 10^{17} > 4 \times 10^{14}$.

$3 \times 10^{17}$ is greater than $4 \times 10^{14}$.

Given

The two expressions to compare are $(5^2) \times 3$ and $(5^2)^3$.

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To Find

Which of the two expressions has a greater value.

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Solution

Let's evaluate each expression separately.

Evaluating the first expression: $(5^2) \times 3$

First, calculate the power $5^2$:

$5^2 = 5 \times 5 = 25$

Now, multiply the result by 3:

$(5^2) \times 3 = 25 \times 3 = 75$

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Evaluating the second expression: $(5^2)^3$

Method 1: Calculate inside the parenthesis first.

$5^2 = 25$

Now, calculate $25^3$:

$25^3 = 25 \times 25 \times 25 = 625 \times 25 = 15625$

Method 2: Using the law of exponents $(a^m)^n = a^{m \times n}$.

$(5^2)^3 = 5^{2 \times 3} = 5^6$

$5^6 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 25 \times 25 \times 25 = 625 \times 25 = 15625$

So, $(5^2)^3 = 15625$.

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Comparison:

We compare the values obtained:

Value of $(5^2) \times 3 = 75$

Value of $(5^2)^3 = 15625$

Since $15625 > 75$, we conclude that $(5^2)^3$ is greater than $(5^2) \times 3$.

Therefore, $(5^2)^3$ is greater.

We will use the law of exponents for power of a product: $(a \times b)^m = a^m \times b^m$.

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(i) $(2 \times 3)^5$

Solution:

Here, $a = 2$, $b = 3$, and $m = 5$.

Applying the law $(a \times b)^m = a^m \times b^m$:

$(2 \times 3)^5 = 2^5 \times 3^5$

The exponential form is $2^5 \times 3^5$.

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(ii) $(2a)^4$

Solution:

We can write $2a$ as $2 \times a$.

Here, the base factors are $2$ and $a$, and the exponent is $m = 4$.

Applying the law $(a \times b)^m = a^m \times b^m$:

$(2a)^4 = (2 \times a)^4 = 2^4 \times a^4$

The exponential form is $2^4 a^4$.

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(iii) $(– 4m)^3$

Solution:

We can write $-4m$ as $(-4) \times m$.

Here, the base factors are $-4$ and $m$, and the exponent is $m = 3$.

Applying the law $(a \times b)^m = a^m \times b^m$:

$(-4m)^3 = ((-4) \times m)^3 = (-4)^3 \times m^3$

The exponential form is $(-4)^3 m^3$.

(Note: This can be further simplified to $-64m^3$, but the question asks for exponential form, which typically involves keeping the bases raised to their powers).

Expansion can mean writing out the repeated multiplication or applying the power rule to the numerator and denominator.

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(i) Expand $\left( \frac{3}{5} \right)^4$

Solution 1 (Repeated Multiplication):

The expression means multiplying the base $\frac{3}{5}$ by itself 4 times:

$\left( \frac{3}{5} \right)^4 = \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}$

This can be further expanded as:

$= \frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5}$

Solution 2 (Using Law of Exponents):

Using the law $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$:

$\left( \frac{3}{5} \right)^4 = \frac{3^4}{5^4}$

The expanded form is $\frac{3^4}{5^4}$ or $\frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5}$.

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(ii) Expand $\left( \frac{-4}{7} \right)^5$

Solution 1 (Repeated Multiplication):

The expression means multiplying the base $\frac{-4}{7}$ by itself 5 times:

$\left( \frac{-4}{7} \right)^5 = \left( \frac{-4}{7} \right) \times \left( \frac{-4}{7} \right) \times \left( \frac{-4}{7} \right) \times \left( \frac{-4}{7} \right) \times \left( \frac{-4}{7} \right)$

This can be further expanded as:

$= \frac{(-4) \times (-4) \times (-4) \times (-4) \times (-4)}{7 \times 7 \times 7 \times 7 \times 7}$

Solution 2 (Using Law of Exponents):

Using the law $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$:

$\left( \frac{-4}{7} \right)^5 = \frac{(-4)^5}{7^5}$

The expanded form is $\frac{(-4)^5}{7^5}$ or $\frac{(-4) \times (-4) \times (-4) \times (-4) \times (-4)}{7 \times 7 \times 7 \times 7 \times 7}$.

Given

The expression $8 \times 8 \times 8 \times 8$.

The desired base for the exponential form is 2.

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To Find

The exponential form of $8 \times 8 \times 8 \times 8$ with base 2.

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Solution

First, we express the current base, 8, as a power of the desired base, 2.

We know that $8 = 2 \times 2 \times 2 = 2^3$.

Now, substitute $2^3$ for each 8 in the given expression:

$8 \times 8 \times 8 \times 8 = (2^3) \times (2^3) \times (2^3) \times (2^3)$

Using the law of exponents for multiplying powers with the same base ($a^m \times a^n = a^{m+n}$):

$(2^3) \times (2^3) \times (2^3) \times (2^3) = 2^{3+3+3+3}$

$= 2^{12}$

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Alternate Solution

First, write the given expression in exponential form with base 8:

$8 \times 8 \times 8 \times 8 = 8^4$

Now, express the base 8 as a power of 2: $8 = 2^3$.

Substitute this into the expression $8^4$:

$8^4 = (2^3)^4$

Using the law of exponents for power of a power ($(a^m)^n = a^{m \times n}$):

$(2^3)^4 = 2^{3 \times 4}$

$= 2^{12}$

Therefore, the exponential form for $8 \times 8 \times 8 \times 8$ taking base as 2 is $2^{12}$.

We will use the laws of exponents to simplify the expressions and write the answers in exponential form.

Relevant Laws:

1. $a^m \times a^n = a^{m+n}$ (Product Rule)

2. $\frac{a^m}{a^n} = a^{m-n}$ (Quotient Rule, $a \neq 0$)

3. $(a^m)^n = a^{m \times n}$ (Power of a Power Rule)

4. $a^m \times b^m = (a \times b)^m$ (Power of a Product Rule)

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(i) $\left( \frac{3^7}{3^2} \right) \times 3^5$

Solution:

First, simplify the term inside the parenthesis using the Quotient Rule:

$\frac{3^7}{3^2} = 3^{7-2} = 3^5$

Now, substitute this back into the expression:

$= 3^5 \times 3^5$

Using the Product Rule:

$= 3^{5+5} = 3^{10}$

The simplified exponential form is $3^{10}$.

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(ii) $2^3 \times 2^2 \times 5^5$

Solution:

Combine the powers of 2 using the Product Rule:

$2^3 \times 2^2 = 2^{3+2} = 2^5$

The expression becomes:

$= 2^5 \times 5^5$

Using the Power of a Product Rule ($a^m \times b^m = (ab)^m$):

$= (2 \times 5)^5$

$= 10^5$

The simplified exponential form is $10^5$.

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(iii) $(6^2 \times 6^4) \div 6^3$

Solution:

First, simplify the term inside the parenthesis using the Product Rule:

$6^2 \times 6^4 = 6^{2+4} = 6^6$

The expression becomes:

$= 6^6 \div 6^3$

Using the Quotient Rule:

$= 6^{6-3} = 6^3$

The simplified exponential form is $6^3$.

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(iv) $[(2^2)^3 \times 3^6] \times 5^6$

Solution:

First, simplify $(2^2)^3$ using the Power of a Power Rule:

$(2^2)^3 = 2^{2 \times 3} = 2^6$

Substitute this back into the expression inside the brackets:

$[2^6 \times 3^6] \times 5^6$

Simplify the term inside the brackets using the Power of a Product Rule ($a^m \times b^m = (ab)^m$):

$2^6 \times 3^6 = (2 \times 3)^6 = 6^6$

The expression becomes:

$= 6^6 \times 5^6$

Using the Power of a Product Rule again:

$= (6 \times 5)^6 = 30^6$

The simplified exponential form is $30^6$.

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(v) $8^2 \div 2^3$

Solution:

First, express the base 8 as a power of 2:

$8 = 2^3$

Substitute this into the expression:

$= (2^3)^2 \div 2^3$

Simplify $(2^3)^2$ using the Power of a Power Rule:

$(2^3)^2 = 2^{3 \times 2} = 2^6$

The expression becomes:

$= 2^6 \div 2^3$

Using the Quotient Rule:

$= 2^{6-3} = 2^3$

The simplified exponential form is $2^3$.

(i) Simplify $\frac{12^4 \; × \; 9^3 \; × \; 4}{6^3 \; × \; 8^2 \; × 27}$

Given Expression:

$\frac{12^4 \; × \; 9^3 \; × \; 4}{6^3 \; × \; 8^2 \; × 27}$

Solution:

First, express each base number as a product of its prime factors:

$12 = 2 \times 2 \times 3 = 2^2 \times 3$

$9 = 3 \times 3 = 3^2$

$4 = 2 \times 2 = 2^2$

$6 = 2 \times 3$

$8 = 2 \times 2 \times 2 = 2^3$

$27 = 3 \times 3 \times 3 = 3^3$

Substitute these prime factor forms into the expression:

$\frac{(2^2 \times 3)^4 \; × \; (3^2)^3 \; × \; 2^2}{(2 \times 3)^3 \; × \; (2^3)^2 \; × \; 3^3}$

Apply the power rules $(ab)^m = a^m b^m$ and $(a^m)^n = a^{mn}$:

$\frac{(2^{2 \times 4} \times 3^4) \; × \; (3^{2 \times 3}) \; × \; 2^2}{(2^3 \times 3^3) \; × \; (2^{3 \times 2}) \; × \; 3^3}$

$\frac{(2^8 \times 3^4) \; × \; 3^6 \; × \; 2^2}{2^3 \times 3^3 \; × \; 2^6 \; × \; 3^3}$

Combine powers with the same base in the numerator and denominator using $a^m \times a^n = a^{m+n}$:

$\frac{2^{8+2} \; × \; 3^{4+6}}{2^{3+6} \; × \; 3^{3+3}}$

$\frac{2^{10} \; × \; 3^{10}}{2^9 \; × \; 3^6}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 2^{10-9} \; × \; 3^{10-6}$

$= 2^1 \; × \; 3^4$

$= 2 \; × \; (3 \times 3 \times 3 \times 3)$

$= 2 \; × \; 81$

$= 162$

The simplified value is 162.

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(ii) Simplify $2^3 × a^3 × 5a^4$

Given Expression:

$2^3 × a^3 × 5a^4$

Solution:

Evaluate the numerical power:

$2^3 = 2 \times 2 \times 2 = 8$

The expression becomes:

$= 8 \times a^3 \times 5 \times a^4$

Rearrange the terms to group constants and variables:

$= (8 \times 5) \times (a^3 \times a^4)$

Multiply the constants:

$8 \times 5 = 40$

Combine the powers of $a$ using the product rule $a^m \times a^n = a^{m+n}$:

$a^3 \times a^4 = a^{3+4} = a^7$

Combine the results:

$= 40a^7$

The simplified expression is $40a^7$.

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(iii) Simplify $\frac{2 \; × \; 3^4 \; × \; 2^5}{9 \; × \; 4^2}$

Given Expression:

$\frac{2 \; × \; 3^4 \; × \; 2^5}{9 \; × \; 4^2}$

Solution:

Express the bases 9 and 4 as powers of their prime factors:

$9 = 3^2$

$4 = 2^2$

Substitute these into the expression:

$\frac{2^1 \; × \; 3^4 \; × \; 2^5}{3^2 \; × \; (2^2)^2}$

Combine powers with the same base in the numerator using $a^m \times a^n = a^{m+n}$:

Numerator: $2^{1+5} \times 3^4 = 2^6 \times 3^4$

Simplify the denominator using $(a^m)^n = a^{mn}$:

Denominator: $3^2 \times 2^{2 \times 2} = 3^2 \times 2^4$

The expression becomes:

$\frac{2^6 \; × \; 3^4}{2^4 \; × \; 3^2}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$= 2^{6-4} \; × \; 3^{4-2}$

$= 2^2 \; × \; 3^2$

Calculate the values:

$= 4 \; × \; 9$

$= 36$

The simplified value is 36.

(Alternatively, the result can be written as $(2 \times 3)^2 = 6^2 = 36$).

(i) $3^2 \times 3^4 \times 3^8$

Solution:

Using the Product Rule for exponents ($a^m \times a^n = a^{m+n}$):

$3^2 \times 3^4 \times 3^8 = 3^{2+4+8}$

$= 3^{14}$

The exponential form is $3^{14}$.

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(ii) $6^{15} \div 6^{10}$

Solution:

Using the Quotient Rule for exponents ($a^m \div a^n = a^{m-n}$):

$6^{15} \div 6^{10} = 6^{15-10}$

$= 6^5$

The exponential form is $6^5$.

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(iii) $a^3 \times a^2$

Solution:

Using the Product Rule for exponents ($a^m \times a^n = a^{m+n}$):

$a^3 \times a^2 = a^{3+2}$

$= a^5$

The exponential form is $a^5$.

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(iv) $7^x \times 7^2$

Solution:

Using the Product Rule for exponents ($a^m \times a^n = a^{m+n}$):

$7^x \times 7^2 = 7^{x+2}$

The exponential form is $7^{x+2}$.

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(v) $(5^2)^3 \div 5^3$

Solution:

First, apply the Power of a Power Rule ($(a^m)^n = a^{m \times n}$):

$(5^2)^3 = 5^{2 \times 3} = 5^6$

The expression becomes $5^6 \div 5^3$.

Now, apply the Quotient Rule ($a^m \div a^n = a^{m-n}$):

$5^6 \div 5^3 = 5^{6-3}$

$= 5^3$

The exponential form is $5^3$.

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(vi) $2^5 \times 5^5$

Solution:

Using the Power of a Product Rule ($a^m \times b^m = (a \times b)^m$):

$2^5 \times 5^5 = (2 \times 5)^5$

$= 10^5$

The exponential form is $10^5$.

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(vii) $a^4 \times b^4$

Solution:

Using the Power of a Product Rule ($a^m \times b^m = (a \times b)^m$):

$a^4 \times b^4 = (a \times b)^4$

$= (ab)^4$

The exponential form is $(ab)^4$.

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(viii) $(3^4)^3$

Solution:

Using the Power of a Power Rule ($(a^m)^n = a^{m \times n}$):

$(3^4)^3 = 3^{4 \times 3}$

$= 3^{12}$

The exponential form is $3^{12}$.

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(ix) $(2^{20} \div 2^{15}) \times 2^3$

Solution:

First, simplify the expression inside the parenthesis using the Quotient Rule ($a^m \div a^n = a^{m-n}$):

$2^{20} \div 2^{15} = 2^{20-15} = 2^5$

The expression becomes $2^5 \times 2^3$.

Now, apply the Product Rule ($a^m \times a^n = a^{m+n}$):

$2^5 \times 2^3 = 2^{5+3}$

$= 2^8$

The exponential form is $2^8$.

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(x) $8^t \div 8^2$

Solution:

Using the Quotient Rule ($a^m \div a^n = a^{m-n}$):

$8^t \div 8^2 = 8^{t-2}$

The exponential form is $8^{t-2}$.

(i) $\frac{2^3 \; × \; 3^4 \; × \; 4}{3 \; × \; 32}$

Solution:

Express 4 and 32 as powers of 2: $4 = 2^2$, $32 = 2^5$.

Also, $3 = 3^1$.

Substitute these into the expression:

$\frac{2^3 \; × \; 3^4 \; × \; 2^2}{3^1 \; × \; 2^5}$

Combine powers of 2 in the numerator using $a^m \times a^n = a^{m+n}$:

$\frac{2^{3+2} \; × \; 3^4}{3^1 \; × \; 2^5} = \frac{2^5 \; × \; 3^4}{3^1 \; × \; 2^5}$

Rearrange terms and apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$= \frac{2^5}{2^5} \; × \; \frac{3^4}{3^1}$

$= 2^{5-5} \; × \; 3^{4-1}$

$= 2^0 \; × \; 3^3$

Since $2^0 = 1$:

$= 1 \; × \; 3^3 = 3^3$

The exponential form is $3^3$.

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(ii) $((5^2)^3 \times 5^4) \div 5^7$

Solution:

Simplify $(5^2)^3$ using $(a^m)^n = a^{mn}$:

$(5^2)^3 = 5^{2 \times 3} = 5^6$

The expression inside the parenthesis becomes $5^6 \times 5^4$.

Simplify $5^6 \times 5^4$ using $a^m \times a^n = a^{m+n}$:

$5^6 \times 5^4 = 5^{6+4} = 5^{10}$

The expression becomes $5^{10} \div 5^7$.

Apply the quotient rule $a^m \div a^n = a^{m-n}$:

$5^{10} \div 5^7 = 5^{10-7} = 5^3$

The exponential form is $5^3$.

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(iii) $25^4 \div 5^3$

Solution:

Express 25 as a power of 5: $25 = 5^2$.

Substitute this into the expression:

$(5^2)^4 \div 5^3$

Apply the power of a power rule $(a^m)^n = a^{mn}$:

$5^{2 \times 4} \div 5^3 = 5^8 \div 5^3$

Apply the quotient rule $a^m \div a^n = a^{m-n}$:

$5^{8-3} = 5^5$

The exponential form is $5^5$.

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(iv) $\frac{3 \; × \; 7^2 \; × \; 11^8}{21 \; × \; 11^3}$

Solution:

Express 21 as a product of its prime factors: $21 = 3 \times 7$.

Substitute this into the expression:

$\frac{3^1 \; × \; 7^2 \; × \; 11^8}{(3^1 \times 7^1) \; × \; 11^3}$

Separate the terms by base:

$= \left( \frac{3^1}{3^1} \right) \; × \; \left( \frac{7^2}{7^1} \right) \; × \; \left( \frac{11^8}{11^3} \right)$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 3^{1-1} \; × \; 7^{2-1} \; × \; 11^{8-3}$

$= 3^0 \; × \; 7^1 \; × \; 11^5$

Since $3^0 = 1$:

$= 1 \; × \; 7^1 \; × \; 11^5 = 7 \times 11^5$

The exponential form is $7 \times 11^5$ (or $7^1 \times 11^5$).

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(v) $\frac{3^7}{3^4 \; × \; 3^3}$

Solution:

Simplify the denominator using the product rule $a^m \times a^n = a^{m+n}$:

$3^4 \times 3^3 = 3^{4+3} = 3^7$

The expression becomes $\frac{3^7}{3^7}$.

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$3^{7-7} = 3^0$

Since $a^0 = 1$ (for $a \neq 0$):

$= 1$

The exponential form is $3^0$ (or 1).

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(vi) $2^0 + 3^0 + 4^0$

Solution:

Using the zero exponent rule $a^0 = 1$ (for $a \neq 0$):

$2^0 = 1$, $3^0 = 1$, $4^0 = 1$.

$= 1 + 1 + 1 = 3$

The exponential form is $3^1$ (or 3).

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(vii) $2^0 × 3^0 × 4^0$

Solution:

Using the zero exponent rule $a^0 = 1$:

$= 1 \times 1 \times 1 = 1$

The exponential form is $1^1$ (or 1 or $a^0$ for any base $a \neq 0$).

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(viii) $(3^0 + 2^0) × 5^0$

Solution:

Using the zero exponent rule $a^0 = 1$:

$3^0 = 1$, $2^0 = 1$, $5^0 = 1$.

$= (1 + 1) \times 1$

$= 2 \times 1 = 2$

The exponential form is $2^1$ (or 2).

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(ix) $\frac{2^8 \; × \; a^5}{4^3 \; × \; a^3}$

Solution:

Express 4 as a power of 2: $4 = 2^2$.

Substitute this into the expression:

$\frac{2^8 \; × \; a^5}{(2^2)^3 \; × \; a^3}$

Apply the power of a power rule $(a^m)^n = a^{mn}$ in the denominator:

$\frac{2^8 \; × \; a^5}{2^{2 \times 3} \; × \; a^3} = \frac{2^8 \; × \; a^5}{2^6 \; × \; a^3}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$= 2^{8-6} \; × \; a^{5-3}$

$= 2^2 \; × \; a^2$

Using the rule $a^m \times b^m = (ab)^m$:

$= (2a)^2$

The exponential form is $2^2 a^2$ or $(2a)^2$.

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(x) $\left( \frac{a^5}{a^3} \right) × a^8$

Solution:

Simplify the term inside the parenthesis using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{a^5}{a^3} = a^{5-3} = a^2$

The expression becomes $a^2 \times a^8$.

Apply the product rule $a^m \times a^n = a^{m+n}$:

$a^{2+8} = a^{10}$

The exponential form is $a^{10}$.

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(xi) $\frac{4^5 \; × \; a^8 b^3}{4^5 \; × \; a^5 b^2}$

Solution:

Apply the quotient rule $\frac{x^m}{x^n} = x^{m-n}$ for each base (4, a, b):

$= 4^{5-5} \; × \; a^{8-5} \; × \; b^{3-2}$

$= 4^0 \; × \; a^3 \; × \; b^1$

Since $4^0 = 1$:

$= 1 \; × \; a^3 \; × \; b^1 = a^3 b$

The exponential form is $a^3 b$ (or $a^3 b^1$).

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(xii) $(2^3 × 2)^2$

Solution:

Simplify the expression inside the parenthesis using the product rule $a^m \times a^n = a^{m+n}$ (note $2 = 2^1$):

$2^3 \times 2^1 = 2^{3+1} = 2^4$

The expression becomes $(2^4)^2$.

Apply the power of a power rule $(a^m)^n = a^{mn}$:

$(2^4)^2 = 2^{4 \times 2} = 2^8$

The exponential form is $2^8$.

(i) $10 \times 10^{11} = 100^{11}$

Justification:

Left Hand Side (LHS): $10 \times 10^{11}$

Using the product rule $a^m \times a^n = a^{m+n}$ (where $10 = 10^1$):

LHS $= 10^1 \times 10^{11} = 10^{1+11} = 10^{12}$

Right Hand Side (RHS): $100^{11}$

Express 100 as a power of 10: $100 = 10^2$.

RHS $= (10^2)^{11}$

Using the power of a power rule $(a^m)^n = a^{mn}$:

RHS $= 10^{2 \times 11} = 10^{22}$

Comparing LHS and RHS: $10^{12} \neq 10^{22}$.

Conclusion: The statement is False.

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(ii) $2^3 > 5^2$

Justification:

Calculate the value of $2^3$: $2^3 = 2 \times 2 \times 2 = 8$.

Calculate the value of $5^2$: $5^2 = 5 \times 5 = 25$.

The statement becomes $8 > 25$.

This inequality is not true, as 8 is less than 25.

Conclusion: The statement is False.

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(iii) $2^3 \times 3^2 = 6^5$

Justification:

Left Hand Side (LHS): $2^3 \times 3^2$

$= (2 \times 2 \times 2) \times (3 \times 3)$

$= 8 \times 9 = 72$.

Right Hand Side (RHS): $6^5$

$= 6 \times 6 \times 6 \times 6 \times 6 = 36 \times 36 \times 6 = 1296 \times 6 = 7776$.

Comparing LHS and RHS: $72 \neq 7776$.

(Note: The law $a^m \times b^n$ cannot be simplified into $(ab)^{m+n}$ or any similar form. The law $a^m \times b^m = (ab)^m$ requires exponents to be the same, and $a^m \times a^n = a^{m+n}$ requires bases to be the same.)

Conclusion: The statement is False.

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(iv) $3^0 = (1000)^0$

Justification:

According to the law of exponents, any non-zero number raised to the power of 0 equals 1 ($a^0 = 1$ for $a \neq 0$).

Left Hand Side (LHS): $3^0 = 1$.

Right Hand Side (RHS): $(1000)^0 = 1$.

Comparing LHS and RHS: $1 = 1$.

Conclusion: The statement is True.

(i) $108 \times 192$

Solution:

First, find the prime factorization of 108:

$\begin{array}{c|cc} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$.

Next, find the prime factorization of 192:

$\begin{array}{c|cc} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$.

Now, multiply the prime factorizations:

$108 \times 192 = (2^2 \times 3^3) \times (2^6 \times 3^1)$

Group the factors with the same base:

$= (2^2 \times 2^6) \times (3^3 \times 3^1)$

Use the product rule $a^m \times a^n = a^{m+n}$:

$= 2^{2+6} \times 3^{3+1}$

$= 2^8 \times 3^4$

The exponential form is $2^8 \times 3^4$.

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(ii) 270

Solution:

Find the prime factorization of 270:

$\begin{array}{c|cc} 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization is $270 = 2 \times 3 \times 3 \times 3 \times 5$.

Writing this in exponential form:

$270 = 2^1 \times 3^3 \times 5^1$

The exponential form is $2 \times 3^3 \times 5$.

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(iii) $729 \times 64$

Solution:

First, find the prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

Next, find the prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

Now, multiply the prime factorizations:

$729 \times 64 = 3^6 \times 2^6$.

The exponential form is $2^6 \times 3^6$ (or $(2 \times 3)^6 = 6^6$, but the question asks for product of prime factors).

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(iv) 768

Solution:

Find the prime factorization of 768:

$\begin{array}{c|cc} 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization is $768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$.

Writing this in exponential form:

$768 = 2^8 \times 3^1$

The exponential form is $2^8 \times 3$.

(i) $\frac{(2^5)^2 \; × \; 7^3}{8^3 \; × \; 7}$

Solution:

Apply the power of a power rule $(a^m)^n = a^{mn}$ to $(2^5)^2$:

$(2^5)^2 = 2^{5 \times 2} = 2^{10}$

Express the base 8 as a power of 2: $8 = 2^3$.

Substitute this into $8^3$: $8^3 = (2^3)^3$.

Apply the power of a power rule: $(2^3)^3 = 2^{3 \times 3} = 2^9$.

Also, $7 = 7^1$.

Substitute these simplified terms back into the expression:

$\frac{2^{10} \; × \; 7^3}{2^9 \; × \; 7^1}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$= 2^{10-9} \; × \; 7^{3-1}$

$= 2^1 \; × \; 7^2$

$= 2 \; × \; (7 \times 7)$

$= 2 \; × \; 49 = 98$

The simplified value is 98 (or $2 \times 7^2$).

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(ii) $\frac{25 \; × \; 5^2 \; × \; t^8}{10^3 \; × \; t^4}$ (Assuming $t \neq 0$)

Solution:

Express 25 as a power of 5: $25 = 5^2$.

Express 10 as a product of prime factors: $10 = 2 \times 5$.

Substitute these into the expression:

$\frac{5^2 \; × \; 5^2 \; × \; t^8}{(2 \times 5)^3 \; × \; t^4}$

Combine powers of 5 in the numerator using $a^m \times a^n = a^{m+n}$:

$5^2 \times 5^2 = 5^{2+2} = 5^4$

Apply the power of a product rule $(ab)^m = a^m b^m$ to the denominator:

$(2 \times 5)^3 = 2^3 \times 5^3$

The expression becomes:

$\frac{5^4 \; × \; t^8}{2^3 \; × \; 5^3 \; × \; t^4}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ for bases 5 and t:

$= \frac{5^{4-3} \; × \; t^{8-4}}{2^3}$

$= \frac{5^1 \; × \; t^4}{2^3}$

$= \frac{5 t^4}{8}$

The simplified expression is $\frac{5t^4}{8}$.

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(iii) $\frac{3^5 \; × \; 10^5 \; × \; 25}{5^7 \; × \; 6^5}$

Solution:

Express the composite bases (10, 25, 6) as products of prime factors:

$10 = 2 \times 5$

$25 = 5^2$

$6 = 2 \times 3$

Substitute these into the expression:

$\frac{3^5 \; × \; (2 \times 5)^5 \; × \; 5^2}{5^7 \; × \; (2 \times 3)^5}$

Apply the power of a product rule $(ab)^m = a^m b^m$:

$\frac{3^5 \; × \; (2^5 \times 5^5) \; × \; 5^2}{5^7 \; × \; (2^5 \times 3^5)}$

Combine powers with the same base in the numerator using $a^m \times a^n = a^{m+n}$:

Numerator: $3^5 \times 2^5 \times 5^{5+2} = 3^5 \times 2^5 \times 5^7$

The expression becomes:

$\frac{2^5 \; × \; 3^5 \; × \; 5^7}{2^5 \; × \; 3^5 \; × \; 5^7}$

Apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$= 2^{5-5} \; × \; 3^{5-5} \; × \; 5^{7-7}$

$= 2^0 \; × \; 3^0 \; × \; 5^0$

Since $a^0 = 1$ (for $a \neq 0$):

$= 1 \times 1 \times 1 = 1$

The simplified value is 1.

Standard form (or scientific notation) expresses a number as a product $a \times 10^n$, where $1 \le a < 10$ and $n$ is an integer.

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(i) 5985.3

Solution:

To get a number between 1 and 10, we need to move the decimal point 3 places to the left:

$5 \underbrace{985.3}_{3 \text{ places}}$ becomes $5.9853$.

Since we moved the decimal 3 places to the left, the power of 10 is 3.

So, $5985.3 = 5.9853 \times 10^3$.

The standard form is $5.9853 \times 10^3$.

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(ii) 65,950

Solution:

The decimal point is assumed to be at the end: $65950.$

To get a number between 1 and 10, we move the decimal point 4 places to the left:

$6 \underbrace{5950.}_{4 \text{ places}}$ becomes $6.5950$ or $6.595$.

Since we moved the decimal 4 places to the left, the power of 10 is 4.

So, $65,950 = 6.595 \times 10^4$.

The standard form is $6.595 \times 10^4$.

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(iii) 3,430,000

Solution:

The decimal point is assumed to be at the end: $3430000.$

To get a number between 1 and 10, we move the decimal point 6 places to the left:

$3 \underbrace{,430,000.}_{6 \text{ places}}$ becomes $3.430000$ or $3.43$.

Since we moved the decimal 6 places to the left, the power of 10 is 6.

So, $3,430,000 = 3.43 \times 10^6$.

The standard form is $3.43 \times 10^6$.

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(iv) 70,040,000,000

Solution:

The decimal point is assumed to be at the end: $70040000000.$

To get a number between 1 and 10, we move the decimal point 10 places to the left:

$7 \underbrace{0,040,000,000.}_{10 \text{ places}}$ becomes $7.0040000000$ or $7.004$.

Since we moved the decimal 10 places to the left, the power of 10 is 10.

So, $70,040,000,000 = 7.004 \times 10^{10}$.

The standard form is $7.004 \times 10^{10}$.

1. 279404

$279404 = 2 \times 100000 + 7 \times 10000 + 9 \times 1000 + 4 \times 100 + 0 \times 10 \ $$ + 4 \times 1$

$= 2 \times 10^5 + 7 \times 10^4 + 9 \times 10^3 + 4 \times 10^2 + 0 \times 10^1 + 4 \times 10^0$

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2. 3006194

$3006194 = 3 \times 1000000 + 0 \times 100000 + 0 \times 10000 + 6 \times 1000 \ $$ + 1 \times 100 \ $$ + 9 \times 10 + 4 \times 1$

$= 3 \times 10^6 + 0 \times 10^5 + 0 \times 10^4 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 \ $$ + 4 \times 10^0$

(Or, omitting zero terms: $3 \times 10^6 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 + 4 \times 10^0$)

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3. 2806196

$2806196 = 2 \times 1000000 + 8 \times 100000 + 0 \times 10000 + 6 \times 1000 \ $$ + 1 \times 100 \ $$ + 9 \times 10 + 6 \times 1$

$= 2 \times 10^6 + 8 \times 10^5 + 0 \times 10^4 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 \ $$ + 6 \times 10^0$

(Or, omitting zero terms: $2 \times 10^6 + 8 \times 10^5 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 + 6 \times 10^0$)

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4. 120719

$120719 = 1 \times 100000 + 2 \times 10000 + 0 \times 1000 + 7 \times 100 + 1 \times 10 \ $$ + 9 \times 1$

$= 1 \times 10^5 + 2 \times 10^4 + 0 \times 10^3 + 7 \times 10^2 + 1 \times 10^1 + 9 \times 10^0$

(Or, omitting zero terms: $1 \times 10^5 + 2 \times 10^4 + 7 \times 10^2 + 1 \times 10^1 + 9 \times 10^0$)

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5. 20068

$20068 = 2 \times 10000 + 0 \times 1000 + 0 \times 100 + 6 \times 10 + 8 \times 1$

$= 2 \times 10^4 + 0 \times 10^3 + 0 \times 10^2 + 6 \times 10^1 + 8 \times 10^0$

(Or, omitting zero terms: $2 \times 10^4 + 6 \times 10^1 + 8 \times 10^0$)

(a) $8 \times 10^4 + 6 \times 10^3 + 0 \times 10^2 + 4 \times 10^1 + 5 \times 10^0$

Solution:

$= 8 \times 10000 + 6 \times 1000 + 0 \times 100 + 4 \times 10 + 5 \times 1$

$= 80000 + 6000 + 0 + 40 + 5$

$= 86045$

The number is 86045.

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(b) $4 \times 10^5 + 5 \times 10^3 + 3 \times 10^2 + 2 \times 10^0$

Solution:

Note that the terms for $10^4$ and $10^1$ are missing, which means their coefficients are 0.

$= 4 \times 10^5 + 0 \times 10^4 + 5 \times 10^3 + 3 \times 10^2 + 0 \times 10^1 + 2 \times 10^0$

$= 4 \times 100000 + 0 \times 10000 + 5 \times 1000 + 3 \times 100 + 0 \times 10 + 2 \times 1$

$= 400000 + 0 + 5000 + 300 + 0 + 2$

$= 405302$

The number is 405302.

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(c) $3 \times 10^4 + 7 \times 10^2 + 5 \times 10^0$

Solution:

Note that the terms for $10^3$ and $10^1$ are missing, which means their coefficients are 0.

$= 3 \times 10^4 + 0 \times 10^3 + 7 \times 10^2 + 0 \times 10^1 + 5 \times 10^0$

$= 3 \times 10000 + 0 \times 1000 + 7 \times 100 + 0 \times 10 + 5 \times 1$

$= 30000 + 0 + 700 + 0 + 5$

$= 30705$

The number is 30705.

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(d) $9 \times 10^5 + 2 \times 10^2 + 3 \times 10^1$

Solution:

Note that the terms for $10^4$, $10^3$, and $10^0$ are missing, which means their coefficients are 0.

$= 9 \times 10^5 + 0 \times 10^4 + 0 \times 10^3 + 2 \times 10^2 + 3 \times 10^1 + 0 \times 10^0$

$= 9 \times 100000 + 0 \times 10000 + 0 \times 1000 + 2 \times 100 + 3 \times 10 + 0 \times 1$

$= 900000 + 0 + 0 + 200 + 30 + 0$

$= 900230$

The number is 900230.

Standard form expresses a number as $a \times 10^n$, where $1 \le a < 10$ and $n$ is an integer.

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(i) 5,00,00,000 (which is 50,000,000)

Solution:

Move the decimal point 7 places to the left to get $5$.

$50,000,000 = 5.0 \times 10^7$

The standard form is $5 \times 10^7$.

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(ii) 70,00,000 (which is 7,000,000)

Solution:

Move the decimal point 6 places to the left to get $7$.

$7,000,000 = 7.0 \times 10^6$

The standard form is $7 \times 10^6$.

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(iii) 3,18,65,00,000 (which is 3,186,500,000)

Solution:

Move the decimal point 9 places to the left to get $3.1865$.

$3,186,500,000 = 3.1865 \times 10^9$

The standard form is $3.1865 \times 10^9$.

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(iv) 3,90,878 (which is 390,878)

Solution:

Move the decimal point 5 places to the left to get $3.90878$.

$390,878 = 3.90878 \times 10^5$

The standard form is $3.90878 \times 10^5$.

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(v) 39087.8

Solution:

Move the decimal point 4 places to the left to get $3.90878$.

$39087.8 = 3.90878 \times 10^4$

The standard form is $3.90878 \times 10^4$.

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(vi) 3908.78

Solution:

Move the decimal point 3 places to the left to get $3.90878$.

$3908.78 = 3.90878 \times 10^3$

The standard form is $3.90878 \times 10^3$.

(a) The distance between Earth and Moon is 384,000,000 m.

Solution:

Move the decimal point 8 places to the left: $3.84$.

Standard form: $3.84 \times 10^8$ m.

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(b) Speed of light in vacuum is 300,000,000 m/s.

Solution:

Move the decimal point 8 places to the left: $3.0$.

Standard form: $3 \times 10^8$ m/s.

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(c) Diameter of the Earth is 1,27,56,000 m (12,756,000 m).

Solution:

Move the decimal point 7 places to the left: $1.2756$.

Standard form: $1.2756 \times 10^7$ m.

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(d) Diameter of the Sun is 1,400,000,000 m.

Solution:

Move the decimal point 9 places to the left: $1.4$.

Standard form: $1.4 \times 10^9$ m.

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(e) In a galaxy there are on an average 100,000,000,000 stars.

Solution:

Move the decimal point 11 places to the left: $1.0$.

Standard form: $1 \times 10^{11}$ stars.

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(f) The universe is estimated to be about 12,000,000,000 years old.

Solution:

Move the decimal point 10 places to the left: $1.2$.

Standard form: $1.2 \times 10^{10}$ years.

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(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.

Solution:

Move the decimal point 20 places to the left: $3.0$.

Standard form: $3 \times 10^{20}$ m.

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(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

Solution:

Move the decimal point 22 places to the left: $6.023$.

Standard form: $6.023 \times 10^{22}$ molecules.

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(i) The earth has 1,353,000,000 cubic km of sea water.

Solution:

Move the decimal point 9 places to the left: $1.353$.

Standard form: $1.353 \times 10^9$ cubic km.

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(j) The population of India was about 1,027,000,000 in March, 2001.

Solution:

Move the decimal point 9 places to the left: $1.027$.

Standard form: $1.027 \times 10^9$.