Chapter 11 Exponents and Powers

To express $256$ as a power of $2$, we perform the prime factorization of $256$ using the base $2$.

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Prime factorization of $256$:

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

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From the prime factorization, we can see that $256$ is the product of eight factors of $2$.

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

$256 = 2^8$

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Therefore, $256$ expressed as a power of $2$ is $2^8$.

We need to compare the values of $2^3$ and $3^2$.

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Calculate the value of $2^3$:

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

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Calculate the value of $3^2$:

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

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Comparing the values, we have $8$ and $9$.

Since $9 > 8$, we conclude that $3^2$ is greater than $2^3$.

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

We need to compare the values of $8^2$ and $2^8$.

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Calculate the value of $8^2$:

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

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Calculate the value of $2^8$:

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

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

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

$2^8 = 16 \times 2 \times 2 \times 2 \times 2$

$2^8 = 32 \times 2 \times 2 \times 2$

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

$2^8 = 128 \times 2$

$2^8 = 256$

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Comparing the values, we have $64$ and $256$.

Since $256 > 64$, we conclude that $2^8$ is greater than $8^2$.

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

We need to expand the given expressions and check if they are all the same.

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Expanding the first expression, $a^3b^2$:

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

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Expanding the second expression, $a^2b^3$:

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

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Expanding the third expression, $b^2a^3$:

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

Using the commutative property of multiplication, we can rearrange the factors:

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

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Expanding the fourth expression, $b^3a^2$:

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

Using the commutative property of multiplication, we can rearrange the factors:

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

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Comparing the expanded forms:

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

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

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

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

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We observe that $a^3b^2$ has three factors of $a$ and two factors of $b$. The expanded form is the same as $b^2a^3$. Thus, $a^3b^2 = b^2a^3$.

We also observe that $a^2b^3$ has two factors of $a$ and three factors of $b$. The expanded form is the same as $b^3a^2$. Thus, $a^2b^3 = b^3a^2$.

However, $a^3b^2$ (three factors of $a$, two of $b$) is different from $a^2b^3$ (two factors of $a$, three of $b$).

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Therefore, the expressions are not all the same. Specifically, $a^3b^2 = b^2a^3$ and $a^2b^3 = b^3a^2$, but $a^3b^2 \ne a^2b^3$ and $b^2a^3 \ne b^3a^2$ (unless $a, b$ are 0 or 1, or $a=b$).

The answer is: No, they are not all same.

We need to express the given numbers as a product of powers of their prime factors.

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(i) $72$

We find the prime factorization of $72$:

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

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

Expressing in terms of powers of prime factors:

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

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(ii) $432$

We find the prime factorization of $432$:

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

$432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

Expressing in terms of powers of prime factors:

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

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(iii) $1000$

We find the prime factorization of $1000$:

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

$1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

Expressing in terms of powers of prime factors:

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

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(iv) $16000$

We find the prime factorization of $16000$:

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

$16000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$

Expressing in terms of powers of prime factors:

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

We need to calculate the values of the given expressions.

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$(1)^5$

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

$(1)^5 = 1$

Value of the expression is: $1$

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(–1)³

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

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

$(-1)^3 = -1$

Value of the expression is: $-1$

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(–1)⁴

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

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

$(-1)^4 = 1$

Value of the expression is: $1$

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(–10)³

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

$(-10)^3 = (100) \times (-10)$

$(-10)^3 = -1000$

Value of the expression is: $-1000$

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(–5)⁴

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

$(-5)^4 = (25) \times (25)$

$(-5)^4 = 625$

Value of the expression is: $625$

We need to find the value of the given expressions:

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(i) $2^6$

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

$2^6 = 4 \times 2 \times 2 \times 2 \times 2$

$2^6 = 8 \times 2 \times 2 \times 2$

$2^6 = 16 \times 2 \times 2$

$2^6 = 32 \times 2$

$2^6 = 64$

Value of the expression is: $64$

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(ii) $9^3$

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

$9 \times 9 = 81$

$81 \times 9 = 729$

$9^3 = 729$

Value of the expression is: $729$

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(iii) $11^2$

$11^2 = 11 \times 11$

$11^2 = 121$

Value of the expression is: $121$

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(iv) $5^4$

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

$5 \times 5 = 25$

$25 \times 5 = 125$

$125 \times 5 = 625$

$5^4 = 625$

Value of the expression is: $625$

We need to express the following products in exponential form:

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(i) $6 \times 6 \times 6 \times 6$

The base is $6$ and it is multiplied by itself $4$ times.

Exponential form is: $6^4$

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(ii) $t \times t$

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

Exponential form is: $t^2$

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(iii) $b \times b \times b \times b$

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

Exponential form is: $b^4$

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(iv) $5 \times 5 \times 7 \times 7 \times 7$

The base $5$ is multiplied by itself $2$ times.

The base $7$ is multiplied by itself $3$ times.

Exponential form is: $5^2 \times 7^3$

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(v) $2 \times 2 \times a \times a$

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

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

Exponential form is: $2^2 \times a^2$

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(vi) $a \times a \times a \times c \times c \times c \times c \times d$

The base $a$ is multiplied by itself $3$ times.

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

The base $d$ is multiplied by itself $1$ time.

Exponential form is: $a^3 \times c^4 \times d$

We need to express each of the given numbers using exponential notation by finding their prime factorization.

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(i) $512$

We find the prime factorization of $512$:

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

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Exponential notation is: $2^9$

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(ii) $343$

We find the prime factorization of $343$:

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

$343 = 7 \times 7 \times 7$

Exponential notation is: $7^3$

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(iii) $729$

We find the prime factorization of $729$:

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

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

Exponential notation is: $3^6$

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(iv) $3125$

We find the prime factorization of $3125$:

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

$3125 = 5 \times 5 \times 5 \times 5 \times 5$

Exponential notation is: $5^5$

We need to calculate the value of each expression and identify the greater number.

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

Calculate $4^3$:

$4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$

Calculate $3^4$:

$3^4 = 3 \times 3 \times 3 \times 3 = 9 \times 3 \times 3 = 27 \times 3 = 81$

Comparing the values: $64$ and $81$.

Since $81 > 64$, $3^4$ is greater than $4^3$.

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

Calculate $5^3$:

$5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$

Calculate $3^5$:

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 3 \times 3 \times 3 = 27 \times 3 \times 3 = 81 \times 3 = 243$

Comparing the values: $125$ and $243$.

Since $243 > 125$, $3^5$ is greater than $5^3$.

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

Calculate $2^8$:

$2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$

Calculate $8^2$:

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

Comparing the values: $256$ and $64$.

Since $256 > 64$, $2^8$ is greater than $8^2$.

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

Consider $2^{100}$ and $100^2$.

$2^{100} = 2^{10 \times 10} = (2^{10})^{10} = (1024)^{10}$

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

Since $1024 > 100$, raising a larger base to a power equal to or greater than the power of a smaller base usually results in a larger number, especially when the bases are significantly different and powers are large.

$(1024)^{10}$ is significantly larger than $100^2 = 10000$. Even $(1024)^2 = 1024 \times 1024$ is greater than $10000$.

Since $(1024)^2 > 100^2$, and $2^{100} = (1024)^{10}$, which is $(1024)^2 \times (1024)^8$, $2^{100}$ is much larger than $100^2$.

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

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

Calculate $2^{10}$:

$2^{10} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 1024$

Calculate $10^2$:

$10^2 = 10 \times 10 = 100$

Comparing the values: $1024$ and $100$.

Since $1024 > 100$, $2^{10}$ is greater than $10^2$.

We need to express each of the given numbers as a product of powers of their prime factors by finding their prime factorization.

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(i) $648$

We find the prime factorization of $648$:

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

$648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

Expressing in terms of powers of prime factors:

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

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(ii) $405$

We find the prime factorization of $405$:

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

$405 = 3 \times 3 \times 3 \times 3 \times 5$

Expressing in terms of powers of prime factors:

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

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(iii) $540$

We find the prime factorization of $540$:

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

$540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5$

Expressing in terms of powers of prime factors:

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

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(iv) $3600$

We find the prime factorization of $3600$:

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

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

Expressing in terms of powers of prime factors:

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

We need to simplify the given expressions.

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(i) $2 \times 10^3$

Calculate the value of $10^3$:

$10^3 = 10 \times 10 \times 10 = 1000$

Now, perform the multiplication:

$2 \times 1000 = 2000$

The simplified value is: $2000$

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

Calculate the value of $7^2$:

$7^2 = 7 \times 7 = 49$

Calculate the value of $2^2$:

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

Now, perform the multiplication:

$49 \times 4 = 196$

The simplified value is: $196$

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(iii) $2^3 \times 5$

Calculate the value of $2^3$:

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

Now, perform the multiplication:

$8 \times 5 = 40$

The simplified value is: $40$

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(iv) $3 \times 4^4$

Calculate the value of $4^4$:

$4^4 = 4 \times 4 \times 4 \times 4 = 16 \times 4 \times 4 = 64 \times 4 = 256$

Now, perform the multiplication:

$3 \times 256 = 768$

The simplified value is: $768$

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(v) $0 \times 10^2$

Calculate the value of $10^2$:

$10^2 = 10 \times 10 = 100$

Now, perform the multiplication:

$0 \times 100 = 0$

The simplified value is: $0$

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

Calculate the value of $5^2$:

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

Calculate the value of $3^3$:

$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

Now, perform the multiplication:

$25 \times 27 = 675$

The simplified value is: $675$

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

Calculate the value of $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 4 \times 2 \times 2 = 8 \times 2 = 16$

Calculate the value of $3^2$:

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

Now, perform the multiplication:

$16 \times 9 = 144$

The simplified value is: $144$

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

Calculate the value of $3^2$:

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

Calculate the value of $10^4$:

$10^4 = 10 \times 10 \times 10 \times 10 = 10000$

Now, perform the multiplication:

$9 \times 10000 = 90000$

The simplified value is: $90000$

We need to simplify the given expressions.

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(i) $(-4)^3$

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

$(-4)^3 = (16) \times (-4)$

$(-4)^3 = -64$

The simplified value is: $-64$

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

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

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

Now, perform the multiplication:

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

Since the product of two negative numbers is positive:

$(-3) \times (-8) = 24$

The simplified value is: $24$

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(iii) $(-3)^2 \times (-5)^2$

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

$(-3)^2 = (-3) \times (-3) = 9$

Next, calculate $(-5)^2$:

$(-5)^2 = (-5) \times (-5) = 25$

Now, perform the multiplication:

$9 \times 25$

$9 \times 25 = 225$

The simplified value is: $225$

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(iv) $(-2)^3 \times (-10)^3$

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

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

Next, calculate $(-10)^3$:

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

Now, perform the multiplication:

$(-8) \times (-1000)$

Since the product of two negative numbers is positive:

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

The simplified value is: $8000$

We need to compare the given pairs of numbers.

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

The first number is $2.7 \times 10^{12}$. The power of $10$ is $12$.

The second number is $1.5 \times 10^{8}$. The power of $10$ is $8$.

Comparing the powers of $10$, we have $12$ and $8$.

Since $12 > 8$, the number with the higher power of $10$ is greater.

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

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

The first number is $4 \times 10^{14}$. The power of $10$ is $14$.

The second number is $3 \times 10^{17}$. The power of $10$ is $17$.

Comparing the powers of $10$, we have $14$ and $17$.

Since $17 > 14$, the number with the higher power of $10$ is greater.

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

We need to compare the values of $(5^2) \times 3$ and $(5^2)^3$.

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Calculate the value of $(5^2) \times 3$:

First, calculate $5^2$:

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

Now, multiply the result by $3$:

$25 \times 3 = 75$

So, $(5^2) \times 3 = 75$.

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Calculate the value of $(5^2)^3$:

Using the law of exponents $(a^m)^n = a^{m \times n}$, we have:

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

Now, calculate the value of $5^6$:

$5^6 = 5 \times 5 \times 5 \times 5 \times 5 \times 5$

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

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

$5^6 = 625 \times 25$

$5^6 = 15625$

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

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Comparing the two values:

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

$(5^2)^3 = 15625$

Since $15625 > 75$, the second expression is greater.

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

We use the law of exponents $(ab)^m = a^m b^m$ to express the given terms in exponential form.

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

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

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

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

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(ii) $(2a)^4$

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

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

We can also calculate $2^4 = 16$.

So the exponential form is: $2^4 \times a^4$ or $16 a^4$.

The expression showing powers of the original factors is preferred for "exponential form": $2^4 \times a^4$

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(iii) $(– 4m)^3$

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

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

We can also calculate $(-4)^3 = (-4) \times (-4) \times (-4) = 16 \times (-4) = -64$.

So the exponential form is: $(-4)^3 \times m^3$ or $-64 m^3$.

The expression showing powers of the original factors is preferred for "exponential form": $(-4)^3 \times m^3$

We need to expand the given expressions using the law of exponents $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$.

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

Apply the exponent to both the numerator and the denominator:

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

Calculate the value of the numerator $3^4$:

$3^4 = 3 \times 3 \times 3 \times 3 = 81$

Calculate the value of the denominator $5^4$:

$5^4 = 5 \times 5 \times 5 \times 5 = 625$

Combine the results:

$\left( \frac{3}{5} \right)^4 = \frac{81}{625}$

The expanded value is: $\frac{81}{625}$

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

Apply the exponent to both the numerator and the denominator:

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

Calculate the value of the numerator $(-4)^5$:

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

$(-4)^5 = 16 \times (-4) \times (-4) \times (-4)$

$(-4)^5 = -64 \times (-4) \times (-4)$

$(-4)^5 = 256 \times (-4)$

$(-4)^5 = -1024$

Calculate the value of the denominator $7^5$:

$7^5 = 7 \times 7 \times 7 \times 7 \times 7$

$7^5 = 49 \times 7 \times 7 \times 7$

$7^5 = 343 \times 7 \times 7$

$7^5 = 2401 \times 7$

$7^5 = 16807$

Combine the results:

$\left( \frac{-4}{7} \right)^5 = \frac{-1024}{16807}$

The expanded value is: $\frac{-1024}{16807}$

The given expression is $8 \times 8 \times 8 \times 8$.

We need to express this in exponential form with base $2$.

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

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

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

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

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

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

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Calculate the sum of the exponents:

$3 + 3 + 3 + 3 = 12$

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So, the exponential form with base $2$ is $2^{12}$.

The exponential form is: $2^{12}$.

We need to simplify the expressions and write the answer in the exponential form.

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

Using the law $\frac{a^m}{a^n} = a^{m-n}$, we simplify the expression inside the parenthesis:

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

Now, multiply the result by $3^5$ using the law $a^m \times a^n = a^{m+n}$:

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

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

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(ii) $2^3 × 2^2 × 5^5$

Using the law $a^m \times a^n = a^{m+n}$, we combine the terms with base $2$:

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

The term with base $5$ remains as is because the base is different.

Combining the results:

$2^5 \times 5^5$

This can also be written using the law $a^m \times b^m = (ab)^m$:

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

The simplified exponential form is: $2^5 \times 5^5$ or $10^5$.

The expression showing powers of the original prime factors is usually preferred: $2^5 \times 5^5$

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(iii) $(6^2 × 6^4) ÷ 6^3$

Using the law $a^m \times a^n = a^{m+n}$, we simplify the expression inside the parenthesis:

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

Now, perform the division using the law $\frac{a^m}{a^n} = a^{m-n}$:

$6^6 ÷ 6^3 = \frac{6^6}{6^3} = 6^{6-3} = 6^3$

The simplified exponential form is: $6^3$

---

(iv) $[(2^2)^3 × 3^6] × 5^6$

Using the law $(a^m)^n = a^{m \times n}$, we simplify $(2^2)^3$:

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

The expression becomes:

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

Using the law $a^m \times b^m = (ab)^m$, we simplify the expression inside the brackets:

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

The expression becomes:

$6^6 \times 5^6$

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

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

The simplified exponential form is: $30^6$.

Alternatively, keeping prime factors separate: $2^6 \times 3^6 \times 5^6$

---

(v) $8^2 ÷ 2^3$

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

Substitute this into the expression:

$(2^3)^2 ÷ 2^3$

Using the law $(a^m)^n = a^{m \times n}$, we simplify $(2^3)^2$:

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

The expression becomes:

$2^6 ÷ 2^3$

Now, perform the division using the law $\frac{a^m}{a^n} = a^{m-n}$:

$2^6 ÷ 2^3 = \frac{2^6}{2^3} = 2^{6-3} = 2^3$

The simplified exponential form is: $2^3$

Solution:

---

We need to simplify the given expressions by using the laws of exponents.

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

First, express all bases in terms of their prime factors:

$12 = 2^2 \times 3$

$9 = 3^2$

$4 = 2^2$

$6 = 2 \times 3$

$8 = 2^3$

$27 = 3^3$

Substitute these into the expression:

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

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

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

$\frac{2^8 \times 3^4 \times 3^6 \times 2^2}{2^3 \times 3^3 \times 2^6 \times 3^3}$

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

Numerator: $2^8 \times 2^2 \times 3^4 \times 3^6 = 2^{8+2} \times 3^{4+6} = 2^{10} \times 3^{10}$

Denominator: $2^3 \times 2^6 \times 3^3 \times 3^3 = 2^{3+6} \times 3^{3+3} = 2^9 \times 3^6$

The expression becomes:

$\frac{2^{10} \times 3^{10}}{2^9 \times 3^6}$

Now, use the division rule $\frac{a^m}{a^n} = a^{m-n}$:

$2^{10-9} \times 3^{10-6}$

$2^1 \times 3^4$

Calculate the value:

$2 \times 3^4 = 2 \times (3 \times 3 \times 3 \times 3) = 2 \times 81 = 162$.

The simplified value is: 162.

---

(ii) $2^3 × a^3 × 5a^4$

Group the terms with the same base:

$(2^3) \times (5) \times (a^3 \times a^4)$

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

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

Combine the numerical coefficients and the variable term:

$2^3 \times 5 \times a^7$

The simplified expression is: $2^3 \times 5 \times a^7$ or $40 a^7$.

---

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

First, express the bases in the denominator in terms of their prime factors:

$9 = 3^2$

$4 = 2^2$

Substitute these into the expression:

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

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

$\frac{2^1 \times 3^4 \times 2^5}{3^2 \times 2^{2 \times 2}}$

$\frac{2^1 \times 3^4 \times 2^5}{3^2 \times 2^4}$

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

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

Rearrange the terms in the denominator for clarity:

Denominator: $2^4 \times 3^2$

The expression becomes:

$\frac{2^6 \times 3^4}{2^4 \times 3^2}$

Now, use the division rule $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$2^{6-4} \times 3^{4-2}$

$2^2 \times 3^2$

Calculate the value:

$2^2 \times 3^2 = (2 \times 2) \times (3 \times 3) = 4 \times 9 = 36$.

The simplified value is: 36.

We use the laws of exponents to simplify the given expressions and write the answer in exponential form.

---

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

Using the law $a^m \times a^n = a^{m+n}$:

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

$3^{2+4+8} = 3^{14}$

The simplified exponential form is: $3^{14}$

---

(ii) $6^{15} ÷ 6^{10}$

Using the law $\frac{a^m}{a^n} = a^{m-n}$:

$6^{15} ÷ 6^{10} = 6^{15-10}$

$6^{15-10} = 6^5$

The simplified exponential form is: $6^5$

---

(iii) $a^3 × a^2$

Using the law $a^m \times a^n = a^{m+n}$:

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

$a^{3+2} = a^5$

The simplified exponential form is: $a^5$

---

(iv) $7^x × 7^2$

Using the law $a^m \times a^n = a^{m+n}$:

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

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

---

(v) $(5^2)^3 ÷ 5^3$

First, using the law $(a^m)^n = a^{mn}$:

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

Now the expression is $5^6 ÷ 5^3$. Using the law $\frac{a^m}{a^n} = a^{m-n}$:

$5^6 ÷ 5^3 = 5^{6-3}$

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

The simplified exponential form is: $5^3$

---

(vi) $2^5 × 5^5$

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

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

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

The simplified exponential form is: $10^5$

---

(vii) $a^4 × b^4$

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

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

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

The simplified exponential form is: $(ab)^4$

---

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

Using the law $(a^m)^n = a^{mn}$:

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

$3^{4 \times 3} = 3^{12}$

The simplified exponential form is: $3^{12}$

---

(ix) $(2^{20} ÷ 2^{15}) × 2^3$

First, simplify the expression in the parenthesis using the law $\frac{a^m}{a^n} = a^{m-n}$:

$2^{20} ÷ 2^{15} = 2^{20-15} = 2^5$

Now the expression is $2^5 \times 2^3$. Using the law $a^m \times a^n = a^{m+n}$:

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

$2^{5+3} = 2^8$

The simplified exponential form is: $2^8$

---

(x) $8^t ÷ 8^2$

Using the law $\frac{a^m}{a^n} = a^{m-n}$:

$8^t ÷ 8^2 = 8^{t-2}$

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

Solution:

---

We use the laws of exponents to simplify the given expressions and express the final answer in exponential form.

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

Express 4 and 32 in terms of prime factors:

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

$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$

Substitute these values into the expression:

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

Group terms with the same base in the numerator and denominator:

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

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

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

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

$2^{5-5} \; × \; 3^{4-1}$

$2^0 \; × \; 3^3$

Since $a^0 = 1$ (for $a \neq 0$), $2^0 = 1$.

$1 \; × \; 3^3 = 3^3$

The simplified expression in exponential form is $3^3$.

---

(ii) ((5²)³ × 5⁴) ÷ 5⁷

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

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

The expression becomes:

$(5^6 \; × \; 5^4) \; ÷ \; 5^7$

Simplify the expression inside the parentheses using the rule $a^m \times a^n = a^{m+n}$:

$5^6 \; × \; 5^4 = 5^{6+4} = 5^{10}$

The expression becomes:

$5^{10} \; ÷ \; 5^7$

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

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

The simplified expression in exponential form is $5^3$.

---

(iii) 25⁴ ÷ 5³

Express the base 25 in terms of its prime factor:

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

Substitute this into the expression:

$(5^2)^4 \; ÷ \; 5^3$

Simplify the term $(5^2)^4$ using the rule $(a^m)^n = a^{mn}$:

$(5^2)^4 = 5^{2 \times 4} = 5^8$

The expression becomes:

$5^8 \; ÷ \; 5^3$

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

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

The simplified expression in exponential form is $5^5$.

---

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

Express the base 21 in terms of its prime factors:

$21 = 3 \times 7$

Substitute this into the expression:

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

Rearrange the terms in the denominator:

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

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

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

$3^0 \; × \; 7^1 \; × \; 11^5$

Since $a^0 = 1$, $3^0 = 1$.

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

The simplified expression in exponential form is $7^1 \; × \; 11^5$ (or $7 \times 11^5$).

---

(v) $\frac{3^7}{3^4 \; × \; 3^3}$

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

$3^4 \; × \; 3^3 = 3^{4+3} = 3^7$

The expression becomes:

$\frac{3^7}{3^7}$

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

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

Since $a^0 = 1$, $3^0 = 1$.

The simplified expression in exponential form is $3^0$ (or simply $1$).

---

(vi) 2⁰ + 3⁰ + 4⁰

Apply the rule $a^0 = 1$ for each term (since the bases are non-zero):

$2^0 = 1$

$3^0 = 1$

$4^0 = 1$

The expression becomes:

$1 + 1 + 1 = 3$

The simplified value is 3. In exponential form, this can be written as $3^1$.

---

(vii) 2⁰ × 3⁰ × 4⁰

Apply the rule $a^0 = 1$ for each term:

$2^0 = 1$

$3^0 = 1$

$4^0 = 1$

The expression becomes:

$1 \times 1 \times 1 = 1$

The simplified value is 1. In exponential form, this can be written as $1$ (or any non-zero base to the power of 0, e.g., $2^0$).

---

(viii) (3⁰ + 2⁰) × 5⁰

Apply the rule $a^0 = 1$ for each term:

$3^0 = 1$

$2^0 = 1$

$5^0 = 1$

The expression becomes:

$(1 + 1) \times 1$

Simplify inside the parentheses:

$2 \times 1 = 2$

The simplified value is 2. In exponential form, this can be written as $2^1$.

---

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

Express the base 4 in the denominator in terms of its prime factor:

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

Substitute this into the expression:

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

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

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

The expression becomes:

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

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

$2^{8-6} \; × \; a^{5-3}$

$2^2 \; × \; a^2$

This can also be written using the rule $a^m \times b^m = (ab)^m$ as $(2a)^2$.

The simplified expression in exponential form is $2^2 \; × \; a^2$ (or $(2a)^2$).

---

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

Simplify the expression inside the parentheses using the rule $\frac{a^m}{a^n} = a^{m-n}$:

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

The expression becomes:

$a^2 \; × \; a^8$

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

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

The simplified expression in exponential form is $a^{10}$.

---

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

Cancel out the common term $4^5$ from the numerator and the denominator:

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

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

$a^{8-5} \; × \; b^{3-2}$

$a^3 \; × \; b^1$

The simplified expression in exponential form is $a^3 \; × \; b^1$ (or $a^3 b$).

---

(xii) (2³ × 2)²

Simplify the expression inside the parentheses using the rule $a^m \times a^n = a^{m+n}$:

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

The expression becomes:

$(2^4)^2$

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

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

The simplified expression in exponential form is $2^8$.

We need to determine if the given statements are true or false and provide justification.

---

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

Consider the left side: $10 \times 10^{11}$

Using the law $a^m \times a^n = a^{m+n}$, where $10 = 10^1$:

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

Consider the right side: $100^{11}$

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

So, $100^{11} = (10^2)^{11}$

Using the law $(a^m)^n = a^{mn}$:

$(10^2)^{11} = 10^{2 \times 11} = 10^{22}$

Comparing the left side ($10^{12}$) and the right side ($10^{22}$), we see that they are not equal since the exponents are different ($12 \ne 22$).

The statement is False.

Justification: $10 \times 10^{11} = 10^{12}$ and $100^{11} = (10^2)^{11} = 10^{22}$. Since $10^{12} \ne 10^{22}$, the statement is false.

---

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

Calculate the value of the left side: $2^3$

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

Calculate the value of the right side: $5^2$

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

Compare the calculated values: $8$ and $25$.

The statement claims $8 > 25$, which is false.

The statement is False.

Justification: $2^3 = 8$ and $5^2 = 25$. Since $8$ is not greater than $25$ ($8 < 25$), the statement is false.

---

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

Calculate the value of the left side: $2^3 \times 3^2$

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

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

$2^3 \times 3^2 = 8 \times 9 = 72$

Calculate the value of the right side: $6^5$

$6^5 = 6 \times 6 \times 6 \times 6 \times 6$

$6^5 = 36 \times 6 \times 6 \times 6$

$6^5 = 216 \times 6 \times 6$

$6^5 = 1296 \times 6$

$6^5 = 7776$

Compare the calculated values: $72$ and $7776$.

The statement claims $72 = 7776$, which is false.

The statement is False.

Justification: $2^3 \times 3^2 = 8 \times 9 = 72$ and $6^5 = 7776$. Since $72 \ne 7776$, the statement is false. Also, the law $a^m \times b^m = (ab)^m$ applies when the exponents are the same, and the law $a^m \times a^n = a^{m+n}$ applies when the bases are the same. Here, neither the bases nor the exponents are the same, so $2^3 \times 3^2 \ne (2 \times 3)^{3+2} = 6^5$.

---

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

Using the law of exponents that states any non-zero number raised to the power of $0$ is equal to $1$ ($a^0 = 1$ for $a \ne 0$).

Calculate the value of the left side: $3^0$

$3^0 = 1$ (since $3 \ne 0$)

Calculate the value of the right side: $(1000)^0$

$(1000)^0 = 1$ (since $1000 \ne 0$)

Compare the calculated values: $1$ and $1$.

The statement claims $1 = 1$, which is true.

The statement is True.

Justification: According to the law of exponents, any non-zero base raised to the power of zero is equal to 1. Therefore, $3^0 = 1$ and $(1000)^0 = 1$, which means $3^0 = (1000)^0$.

We need to express each of the following as a product of prime factors only in exponential form.

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(i) $108 \times 192$

First, find the prime factorization of $108$:

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

$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$

Next, find the prime factorization of $192$:

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

$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$

Now, multiply the exponential forms and combine the powers of the same prime factors:

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

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

Using the law $a^m \times a^n = a^{m+n}$:

$108 \times 192 = 2^{2+6} \times 3^{3+1}$

$108 \times 192 = 2^8 \times 3^4$

The expression as a product of prime factors in exponential form is: $2^8 \times 3^4$

---

(ii) $270$

Find the prime factorization of $270$:

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

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

Expressing in terms of powers of prime factors:

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

The expression as a product of prime factors in exponential form is: $2^1 \times 3^3 \times 5^1$

---

(iii) $729 \times 64$

First, find the prime factorization of $729$:

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

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$

Next, find the prime factorization of $64$:

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

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$

Now, multiply the exponential forms:

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

Using the law $a^m \times b^m = (ab)^m$ (optional, but requested form is product of prime factors):

$3^6 \times 2^6 = (3 \times 2)^6 = 6^6$. While this is a valid exponential form, the question asks for a product of *prime* factors.

The expression as a product of prime factors in exponential form is: $2^6 \times 3^6$

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(iv) $768$

Find the prime factorization of $768$:

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

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

Expressing in terms of powers of prime factors:

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

The expression as a product of prime factors in exponential form is: $2^8 \times 3^1$

Solution:

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We need to simplify the given expressions by using the laws of exponents.

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

Express $8$ as a power of $2$:

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

Substitute this into the expression and apply the power of a power rule $(a^m)^n = a^{mn}$:

$\frac{(2^5)^2 \; × \; 7^3}{(2^3)^3 \; × \; 7^1} = \frac{2^{5 \times 2} \; × \; 7^3}{2^{3 \times 3} \; × \; 7^1} = \frac{2^{10} \; × \; 7^3}{2^9 \; × \; 7^1}$

Use the division law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$2^{10-9} \times 7^{3-1} = 2^1 \times 7^2$

Calculate the value:

$2^1 \times 7^2 = 2 \times (7 \times 7) = 2 \times 49 = 98$.

The simplified value is 98.

---

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

Express $25$ and $10$ in terms of their prime factors:

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

$10 = 2 \times 5$

Substitute these into the expression and apply the power of a product rule $(ab)^m = a^m b^m$:

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

Use the division law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$\frac{5^4}{5^3} \times \frac{t^8}{t^4} \times \frac{1}{2^3} = 5^{4-3} \times t^{8-4} \times 2^{-3} = 5^1 \times t^4 \times \frac{1}{2^3}$

The simplified expression is $\frac{5 \times t^4}{2^3}$ (or $\frac{5t^4}{8}$).

---

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

Express $10$, $25$, and $6$ in terms of their prime factors:

$10 = 2 \times 5$

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

$6 = 2 \times 3$

Substitute these into the expression and apply the power of a product rule $(ab)^m = a^m b^m$:

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

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

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

Denominator: $2^5 \times 3^5 \times 5^7$

The expression becomes:

$\frac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7}$

Use the division law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for each base:

$2^{5-5} \times 3^{5-5} \times 5^{7-7} = 2^0 \times 3^0 \times 5^0$

Using the law $a^0 = 1$ for non-zero bases:

$1 \times 1 \times 1 = 1$

The simplified value is 1.

We need to express the given numbers in standard form, which is $M \times 10^n$, where $1 \leq M < 10$ and $n$ is an integer.

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(i) 5985.3

To write this in standard form, we move the decimal point to the left so that there is only one non-zero digit to its left. The first non-zero digit is $5$.

We move the decimal point $3$ places to the left: $5.9853$.

Since we moved the decimal point $3$ places to the left, the exponent of $10$ is positive $3$.

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

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(ii) 65,950

The number is $65,950$. We can write this as $65950.0$.

To write this in standard form, we move the decimal point to the left so that there is only one non-zero digit to its left. The first non-zero digit is $6$.

We move the decimal point $4$ places to the left: $6.5950$.

Since we moved the decimal point $4$ places to the left, the exponent of $10$ is positive $4$.

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

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(iii) 3,430,000

The number is $3,430,000$. We can write this as $3430000.0$.

To write this in standard form, we move the decimal point to the left so that there is only one non-zero digit to its left. The first non-zero digit is $3$.

We move the decimal point $6$ places to the left: $3.430000$.

Since we moved the decimal point $6$ places to the left, the exponent of $10$ is positive $6$.

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

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(iv) 70,040,000,000

The number is $70,040,000,000$. We can write this as $70040000000.0$.

To write this in standard form, we move the decimal point to the left so that there is only one non-zero digit to its left. The first non-zero digit is $7$.

We move the decimal point $10$ places to the left: $7.0040000000$.

Since we moved the decimal point $10$ places to the left, the exponent of $10$ is positive $10$.

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

For $279404$:

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

$279404 = 2 \times 10^5 + 7 \times 10^4 + 9 \times 10^3 + 4 \times 10^2 + 0 \times 10^1 + 4 \times 10^0$

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For $3006194$:

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

$3006194 = 3 \times 10^6 + 0 \times 10^5 + 0 \times 10^4 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 + 4 \times 10^0$

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For $2806196$:

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

$2806196 = 2 \times 10^6 + 8 \times 10^5 + 0 \times 10^4 + 6 \times 10^3 + 1 \times 10^2 + 9 \times 10^1 + 6 \times 10^0$

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For $120719$:

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

$120719 = 1 \times 10^5 + 2 \times 10^4 + 0 \times 10^3 + 7 \times 10^2 + 1 \times 10^1 + 9 \times 10^0$

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For $20068$:

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

$20068 = 2 \times 10^4 + 0 \times 10^3 + 0 \times 10^2 + 6 \times 10^1 + 8 \times 10^0$

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

Calculate the value of each term:

$8 \times 10^4 = 8 \times 10000 = 80000$

$6 \times 10^3 = 6 \times 1000 = 6000$

$0 \times 10^2 = 0 \times 100 = 0$

$4 \times 10^1 = 4 \times 10 = 40$

$5 \times 10^0 = 5 \times 1 = 5$

Summing the values:

$80000 + 6000 + 0 + 40 + 5 = 86045$

The number is: $86045$

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

Calculate the value of each term:

$4 \times 10^5 = 4 \times 100000 = 400000$

$5 \times 10^3 = 5 \times 1000 = 5000$

$3 \times 10^2 = 3 \times 100 = 300$

$2 \times 10^0 = 2 \times 1 = 2$

Summing the values (inserting 0 for missing powers of 10):

$400000 + (0 \times 10^4) + 5000 + 300 + (0 \times 10^1) + 2$

$400000 + 0 + 5000 + 300 + 0 + 2 = 405302$

The number is: $405302$

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

Calculate the value of each term:

$3 \times 10^4 = 3 \times 10000 = 30000$

$7 \times 10^2 = 7 \times 100 = 700$

$5 \times 10^0 = 5 \times 1 = 5$

Summing the values (inserting 0 for missing powers of 10):

$30000 + (0 \times 10^3) + 700 + (0 \times 10^1) + 5$

$30000 + 0 + 700 + 0 + 5 = 30705$

The number is: $30705$

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

Calculate the value of each term:

$9 \times 10^5 = 9 \times 100000 = 900000$

$2 \times 10^2 = 2 \times 100 = 200$

$3 \times 10^1 = 3 \times 10 = 30$

Summing the values (inserting 0 for missing powers of 10):

$900000 + (0 \times 10^4) + (0 \times 10^3) + 200 + 30 + (0 \times 10^0)$

$900000 + 0 + 0 + 200 + 30 + 0 = 900230$

The number is: $900230$

Solution:

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We need to express the given numbers in standard form, which is $M \times 10^n$, where $1 \leq M < 10$ and $n$ is an integer. This is also known as scientific notation.

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(i) 5,00,00,000

The number is 5,00,00,000. The first non-zero digit is 5.

To get a number between 1 and 10, we place the decimal after the first digit: 5.

The original number can be thought of as having the decimal at the end (5,00,00,000.). We need to move the decimal point to the left until it is after the first digit 5. Count the number of places the decimal is moved. The original number is 50,000,000.

Moving the decimal 7 places to the left gives 5.0.

The exponent of 10 is 7 (since we moved the decimal left 7 places).

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

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(ii) 70,00,000

The number is 70,00,000. The first non-zero digit is 7.

To get a number between 1 and 10, we place the decimal after the first digit: 7.

The original number is 7,000,000. Moving the decimal 6 places to the left gives 7.0.

The exponent of 10 is 6 (since we moved the decimal left 6 places).

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

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(iii) 3,18,65,00,000

The number is 3,18,65,00,000. The first non-zero digit is 3.

To get a number between 1 and 10, we place the decimal after the first digit: 3.

The original number is 3,186,500,000. Moving the decimal 9 places to the left gives 3.1865.

The exponent of 10 is 9 (since we moved the decimal left 9 places).

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

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(iv) 3,90,878

The number is 3,90,878. The first non-zero digit is 3.

To get a number between 1 and 10, we place the decimal after the first digit: 3.

The original number is 390,878. Moving the decimal 5 places to the left gives 3.90878.

The exponent of 10 is 5 (since we moved the decimal left 5 places).

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

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(v) 39087.8

The number is 39087.8. The first non-zero digit is 3.

To get a number between 1 and 10, we place the decimal after the first digit: 3.

The original decimal point is between 7 and 8. We need to move it to after the first digit 3. Moving the decimal 4 places to the left gives 3.90878.

The exponent of 10 is 4 (since we moved the decimal left 4 places).

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

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(vi) 3908.78

The number is 3908.78. The first non-zero digit is 3.

To get a number between 1 and 10, we place the decimal after the first digit: 3.

The original decimal point is between 8 and 7. We need to move it to after the first digit 3. Moving the decimal 3 places to the left gives 3.90878.

The exponent of 10 is 3 (since we moved the decimal left 3 places).

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

Solution:

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We need to express the numbers appearing in the following statements in standard form ($M \times 10^n$, where $1 \leq M < 10$ and $n$ is an integer).

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

The number is 38,40,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 3, i.e., 3.84.

We moved the decimal 8 places to the left (from the end of 384,000,000).

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

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

The number is 30,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 3, i.e., 3.

We moved the decimal 8 places to the left.

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

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

The number is 1,27,56,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.2756.

We moved the decimal 7 places to the left.

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

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

The number is 1,40,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.4.

We moved the decimal 9 places to the left.

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.

The number is 1,00,00,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.

We moved the decimal 11 places to the left.

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.

The number is 12,00,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.2.

We moved the decimal 10 places to the left.

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.

The number is 3,00,00,00,00,00,00,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 3, i.e., 3.

We moved the decimal 20 places to the left.

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.

The number is 6,02,30,00,00,00,00,00,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 6, i.e., 6.023.

We moved the decimal 22 places to the left.

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.

The number is 1,35,30,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.353.

We moved the decimal 9 places to the left.

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.

The number is 1,02,70,00,000.

To get a number between 1 and 10, we place the decimal after the first digit 1, i.e., 1.027.

We moved the decimal 9 places to the left.

Standard form: $1.027 \times 10^9$