Chapter 12 Exponents and Powers

Concept Used:

The definition of a negative exponent states that for any non-zero base $a$ and any positive integer $m$, $a^{-m} = \frac{1}{a^m}$.

Conversely, $\frac{1}{a^{-m}} = a^m$.

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Solution:

(i) Find the value of $2^{-3}$.

Using the rule $a^{-m} = \frac{1}{a^m}$ with $a=2$ and $m=3$:

$2^{-3} = \frac{1}{2^3}$

Calculate the value of $2^3$:

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

So, $2^{-3} = \frac{1}{8}$.

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(ii) Find the value of $\frac{1}{3^{-2}}$.

Using the rule $\frac{1}{a^{-m}} = a^m$ with $a=3$ and $m=2$:

$\frac{1}{3^{-2}} = 3^2$

Calculate the value of $3^2$:

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

So, $\frac{1}{3^{-2}} = 9$.

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Conclusion:

(i) The value of $2^{-3}$ is $\frac{1}{8}$.

(ii) The value of $\frac{1}{3^{-2}}$ is $9$.

Concept Used:

The laws of exponents state that for any non-zero base $a$ and integers $m$ and $n$:

• $a^m \times a^n = a^{m+n}$ (Product rule)
• $a^m \div a^n = a^{m-n}$ (Quotient rule)

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Solution:

(i) Simplify $(– 4)^5 \times (– 4)^{–10}$.

Here, the base is $a = -4$, and the exponents are $m=5$ and $n=-10$.

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

$(– 4)^5 \times (– 4)^{–10} = (– 4)^{5 + (–10)}$

$(– 4)^5 \times (– 4)^{–10} = (– 4)^{5 - 10}$

$(– 4)^5 \times (– 4)^{–10} = (– 4)^{-5}$

Now, use the definition of a negative exponent ($a^{-m} = \frac{1}{a^m}$):

$(– 4)^{-5} = \frac{1}{(– 4)^5}$

Calculate $(– 4)^5 = (– 4) \times (– 4) \times (– 4) \times (– 4) \times (– 4)$. Since the exponent is odd, the result will be negative.

$(– 4)^5 = - (4 \times 4 \times 4 \times 4 \times 4) = - (16 \times 16 \times 4) = - (256 \times 4) \ $$ = -1024$.

So, $(– 4)^{-5} = \frac{1}{-1024} = -\frac{1}{1024}$.

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(ii) Simplify $2^5 \div 2^{– 6}$.

Here, the base is $a=2$, and the exponents are $m=5$ and $n=-6$.

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

$2^5 \div 2^{– 6} = 2^{5 - (– 6)}$

$2^5 \div 2^{– 6} = 2^{5 + 6}$

$2^5 \div 2^{– 6} = 2^{11}$

Calculate the value of $2^{11}$:

$2^{11} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$2^{11} = 32 \times 32 \times 32 \times 2 = 1024 \times 32 \times 2 = 1024 \times 64$

$1024 \times 64 = 65536$.

So, $2^5 \div 2^{– 6} = 65536$.

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Conclusion:

(i) $(– 4)^5 \times (– 4)^{–10} = -\frac{1}{1024}$.

(ii) $2^5 \div 2^{– 6} = 65536$.

Problem Description:

We need to rewrite the expression $4^{-3}$ so that the base is 2 instead of 4.

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To Express:

$4^{-3}$ as a power with base 2.

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Solution:

We know that the base 4 can be expressed as a power of 2: $4 = 2^2$.

Substitute $4 = 2^2$ into the given expression $4^{-3}$:

$4^{-3} = (2^2)^{-3}$

Now, use the law of exponents that states $(a^m)^n = a^{mn}$. Here, $a=2$, $m=2$, and $n=-3$.

$(2^2)^{-3} = 2^{2 \times (–3)}$

$(2^2)^{-3} = 2^{-6}$.

The expression $4^{-3}$ is now expressed as a power with the base 2.

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Conclusion:

Therefore, $4^{-3}$ can be expressed as $2^{-6}$.

To simplify the given expressions and write the answers in exponential form, we will use the laws of exponents.

Laws of Exponents to be used:

• $a^m \div a^n = a^{m-n}$
• $(a^m)^n = a^{m \times n}$
• $a^m \times a^n = a^{m+n}$
• $a^m \times b^m = (a \times b)^m$
• $a^{-n} = \frac{1}{a^n}$

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(i) (2⁵ ÷ 2⁸)⁵ × 2^–5

First, simplify the expression inside the bracket using the law $a^m \div a^n = a^{m-n}$.

$(2^{5-8})^5 \times 2^{-5}$

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

Now, apply the law $(a^m)^n = a^{m \times n}$.

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

$= 2^{-15} \times 2^{-5}$

Finally, use the law $a^m \times a^n = a^{m+n}$.

$= 2^{-15 + (-5)}$

$= 2^{-15 - 5}$

$= 2^{-20}$

The simplified exponential form is $2^{-20}$.

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(ii) (–4)^–3 × (5)^–3 × (–5)^–3

The exponents are the same for all three terms. We can use the law $a^m \times b^m \times c^m = (a \times b \times c)^m$.

$= ((-4) \times 5 \times (-5))^{-3}$

Multiply the numbers inside the bracket:

$= (-20 \times -5)^{-3}$

$= (100)^{-3}$

The simplified exponential form is $100^{-3}$.

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(iii) $\frac{1}{8}$ × (3)^-3

First, express 8 in exponential form. We know that $8 = 2^3$.

So, $\frac{1}{8} = \frac{1}{2^3}$.

Using the law $\frac{1}{a^n} = a^{-n}$, we can write $\frac{1}{2^3} = 2^{-3}$.

The expression becomes:

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

Now, we have the same exponent. Using the law $a^m \times b^m = (a \times b)^m$:

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

$= 6^{-3}$

The simplified exponential form is $6^{-3}$.

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(iv) (-3)⁴ × $\left( \frac{5}{3} \right)^{4}$

The exponents are the same. We can use the law $a^m \times b^m = (a \times b)^m$.

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

Simplify the expression inside the bracket:

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

$= \left( \frac{\cancel{-3}^{-1} \times 5}{\cancel{3}_1} \right)^4$

$= (-1 \times 5)^4$

$= (-5)^4$

The simplified exponential form is $(-5)^4$.

Given:

The equation $(–3)^{m + 1} \times (–3)^{5} = (–3)^{7}$.

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To Find:

The value of $m$.

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Solution:

We are given the equation:

The left side of the equation involves the multiplication of two terms with the same base ($–3$). We can use the product rule of exponents, which states that $a^m \times a^n = a^{m+n}$.

Applying the product rule to the left side of equation (1):

$(–3)^{m + 1} \times (–3)^{5} = (–3)^{(m + 1) + 5}$

$(–3)^{m + 1} \times (–3)^{5} = (–3)^{m + 6}$.

Now, substitute this back into the original equation (1):

Equation (2) states that two exponential expressions with the same non-zero base are equal. This implies that their exponents must be equal.

Therefore, we can equate the exponents:

$m + 6 = 7$

Solve for $m$ by subtracting 6 from both sides of the equation:

$m = 7 - 6$

$m = 1$.

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Conclusion:

The value of $m$ is $\boxed{1}$.

Problem Description:

We need to evaluate the given expression which has a negative exponent.

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To Find:

The value of $\left( \frac{2}{3} \right)^{-2}$.

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Solution:

We have the expression $\left( \frac{2}{3} \right)^{-2}$.

We can use the law of exponents that states $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$. Here, $a=2$, $b=3$, and $m=2$.

Applying this rule:

$\left( \frac{2}{3} \right)^{-2} = \left( \frac{3}{2} \right)^{2}$

Now, apply the power to both the numerator and the denominator:

$\left( \frac{3}{2} \right)^{2} = \frac{3^2}{2^2}$

Calculate the values of $3^2$ and $2^2$:

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

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

So, $\frac{3^2}{2^2} = \frac{9}{4}$.

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Alternate Method:

We can also use the rule $\left( \frac{a}{b} \right)^{-m} = \frac{a^{-m}}{b^{-m}}$ and $a^{-m} = \frac{1}{a^m}$.

$\left( \frac{2}{3} \right)^{-2} = \frac{2^{-2}}{3^{-2}}$

Using $a^{-m} = \frac{1}{a^m}$:

$\frac{2^{-2}}{3^{-2}} = \frac{\frac{1}{2^2}}{\frac{1}{3^2}}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$\frac{\frac{1}{2^2}}{\frac{1}{3^2}} = \frac{1}{2^2} \times \frac{3^2}{1} = \frac{3^2}{2^2}$

Calculate the values:

$\frac{3^2}{2^2} = \frac{9}{4}$.

Both methods yield the same result.

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Conclusion:

The value of $\left( \frac{2}{3} \right)^{-2}$ is $\frac{9}{4}$.

Concept Used:

Laws of exponents:

• $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$
• $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$
• $a^m \times a^n = a^{m+n}$
• $a^m \times b^m = (ab)^m$

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Solution:

(i) Simplify $\left\{ \left( \frac{1}{3} \right)^{-2} - \left( \frac{1}{2} \right)^{-3}\right\}$ ÷ $\left( \frac{1}{4} \right)^{-2}$.

First, evaluate the terms with negative exponents within the curly braces using the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$:

$\left( \frac{1}{3} \right)^{-2} = \left( \frac{3}{1} \right)^2 = 3^2 = 9$.

$\left( \frac{1}{2} \right)^{-3} = \left( \frac{2}{1} \right)^3 = 2^3 = 8$.

The expression inside the curly braces is now $9 - 8 = 1$.

Next, evaluate the term outside the curly braces:

$\left( \frac{1}{4} \right)^{-2} = \left( \frac{4}{1} \right)^2 = 4^2 = 16$.

The entire expression is now $1$ ÷ $16$.

$1 \div 16 = \frac{1}{16}$.

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(ii) Simplify $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-5}$.

We have two terms multiplied together. Notice that the bases $\frac{5}{8}$ and $\frac{8}{5}$ are reciprocals of each other.

Use the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$ on the first term:

$\left( \frac{5}{8} \right)^{-7} = \left( \frac{8}{5} \right)^{7}$.

Now the expression becomes $\left( \frac{8}{5} \right)^{7} \times \left( \frac{8}{5} \right)^{-5}$.

We have the multiplication of two terms with the same base ($\frac{8}{5}$). Use the product rule of exponents ($a^m \times a^n = a^{m+n}$):

$\left( \frac{8}{5} \right)^{7} \times \left( \frac{8}{5} \right)^{-5} = \left( \frac{8}{5} \right)^{7 + (–5)}$

$\left( \frac{8}{5} \right)^{7 + (–5)} = \left( \frac{8}{5} \right)^{7 - 5} = \left( \frac{8}{5} \right)^{2}$.

Now, apply the power to the numerator and the denominator:

$\left( \frac{8}{5} \right)^{2} = \frac{8^2}{5^2}$

Calculate the values of $8^2$ and $5^2$:

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

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

So, $\frac{8^2}{5^2} = \frac{64}{25}$.

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Conclusion:

(i) $\left\{ \left( \frac{1}{3} \right)^{-2} - \left( \frac{1}{2} \right)^{-3}\right\}$ ÷ $\left( \frac{1}{4} \right)^{-2} = \frac{1}{16}$.

(ii) $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-5} = \frac{64}{25}$.

Concept Used:

The definition of a negative exponent states that for any non-zero base $a$ and any positive integer $m$, $a^{-m} = \frac{1}{a^m}$.

Also, $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m = \frac{b^m}{a^m}$.

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Solution:

(i) Evaluate $3^{-2}$.

Using the rule $a^{-m} = \frac{1}{a^m}$ with $a=3$ and $m=2$:

$3^{-2} = \frac{1}{3^2}$

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

So, $3^{-2} = \frac{1}{9}$.

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(ii) Evaluate $(– 4)^{-2}$.

Using the rule $a^{-m} = \frac{1}{a^m}$ with $a=-4$ and $m=2$:

$(– 4)^{-2} = \frac{1}{(– 4)^2}$

Calculate $(– 4)^2 = (– 4) \times (– 4)$. Since the exponent is even, the result is positive.

$(– 4)^2 = 16$.

So, $(– 4)^{-2} = \frac{1}{16}$.

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(iii) Evaluate $\left( \frac{1}{2} \right)^{-5}$.

Using the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$ with $a=1$, $b=2$, and $m=5$:

$\left( \frac{1}{2} \right)^{-5} = \left( \frac{2}{1} \right)^{5} = 2^5$.

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

So, $\left( \frac{1}{2} \right)^{-5} = 32$.

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Conclusion:

(i) $3^{-2} = \frac{1}{9}$.

(ii) $(– 4)^{-2} = \frac{1}{16}$.

(iii) $\left( \frac{1}{2} \right)^{-5} = 32$.

Given:

The expressions to be simplified.

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To Simplify:

Simplify the given expressions and express the result in power notation with a positive exponent.

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Solution:

(i) Simplify $(– 4)^{5} \div (– 4)^{8}$.

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

$(– 4)^{5} \div (– 4)^{8} = (– 4)^{5 - 8}$

$(– 4)^{5} \div (– 4)^{8} = (– 4)^{-3}$

To express the result with a positive exponent, use the rule $a^{-m} = \frac{1}{a^m}$:

$(– 4)^{-3} = \frac{1}{(– 4)^3}$.

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(ii) Simplify $\left( \frac{1}{2^3} \right)^{2}$.

We can write $\frac{1}{2^3}$ as $2^{-3}$ using the rule $\frac{1}{a^m} = a^{-m}$.

The expression becomes $(2^{-3})^{2}$.

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

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

To express the result with a positive exponent, use the rule $a^{-m} = \frac{1}{a^m}$:

$2^{-6} = \frac{1}{2^6}$.

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(iii) Simplify $(-3)^{4} \times \left( \frac{5}{3} \right)^{4}$.

Here, the exponents are the same. Using the rule $a^m \times b^m = (ab)^m$:

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

Simplify the term inside the parentheses:

$-3 \times \frac{5}{3} = -\cancel{3} \times \frac{5}{\cancel{3}} = -5$.

The expression becomes $(-5)^4$.

Since the base is negative and the exponent is even, $(-5)^4 = 5^4$.

The result is $5^4$. The exponent is already positive.

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(iv) Simplify $(3^{-7} \div 3^{-10}) \times 3^{-5}$.

First, simplify the expression inside the parentheses using the quotient rule $a^m \div a^n = a^{m-n}$:

$3^{-7} \div 3^{-10} = 3^{-7 - (-10)} = 3^{-7 + 10} = 3^3$.

The expression becomes $3^3 \times 3^{-5}$.

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

$3^3 \times 3^{-5} = 3^{3 + (-5)} = 3^{3 - 5} = 3^{-2}$.

To express the result with a positive exponent, use the rule $a^{-m} = \frac{1}{a^m}$:

$3^{-2} = \frac{1}{3^2}$.

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(v) Simplify $2^{-3} \times (–7)^{-3}$.

Here, the exponents are the same. Using the rule $a^m \times b^m = (ab)^m$:

$2^{-3} \times (–7)^{-3} = (2 \times –7)^{-3}$

Simplify the term inside the parentheses:

$2 \times –7 = –14$.

The expression becomes $(–14)^{-3}$.

To express the result with a positive exponent, use the rule $a^{-m} = \frac{1}{a^m}$:

$(–14)^{-3} = \frac{1}{(–14)^3}$.

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Conclusion:

The simplified expressions with positive exponents are:

(i) $\frac{1}{(– 4)^3}$

(ii) $\frac{1}{2^6}$

(iii) $5^4$

(iv) $\frac{1}{3^2}$

(v) $\frac{1}{(–14)^3}$

Concept Used:

Laws of exponents:

• $a^0 = 1$ (for any non-zero base $a$)
• $a^{-m} = \frac{1}{a^m}$
• $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m = \frac{b^m}{a^m}$
• $a^m \times b^m = (ab)^m$
• $(a^m)^n = a^{mn}$

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Solution:

(i) Evaluate $(3^0 + 4^{-1}) \times 2^2$.

Evaluate the terms inside the parentheses:

$3^0 = 1$ (Using $a^0 = 1$).

$4^{-1} = \frac{1}{4^1} = \frac{1}{4}$ (Using $a^{-m} = \frac{1}{a^m}$).

The expression inside the parentheses is $1 + \frac{1}{4}$.

$1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4}$.

Evaluate the term outside the parentheses:

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

Now, multiply the results:

$\left( \frac{5}{4} \right) \times 4 = \frac{5}{\cancel{4}} \times \cancel{4} = 5$.

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(ii) Evaluate $(2^{-1} \times 4^{-1}) \div 2^{-2}$.

Evaluate the terms inside the parentheses:

Using $a^{-m} = \frac{1}{a^m}$:

$2^{-1} = \frac{1}{2}$.

$4^{-1} = \frac{1}{4}$.

The expression inside the parentheses is $\frac{1}{2} \times \frac{1}{4} = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}$.

Alternatively, use the rule $a^m \times b^m = (ab)^m$ for the parentheses:

$2^{-1} \times 4^{-1} = (2 \times 4)^{-1} = 8^{-1} = \frac{1}{8}$.

Evaluate the term outside the parentheses:

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$.

Now, perform the division:

$\frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times \frac{4}{1}$

$\frac{1}{\cancel{8}_{2}} \times \frac{\cancel{4}^{1}}{1} = \frac{1 \times 1}{2 \times 1} = \frac{1}{2}$.

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(iii) Evaluate $\left( \frac{1}{2} \right)^{-2}$ + $\left( \frac{1}{3} \right)^{-2}$ + $\left( \frac{1}{4} \right)^{-2}$.

Use the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$ for each term:

$\left( \frac{1}{2} \right)^{-2} = \left( \frac{2}{1} \right)^{2} = 2^2 = 4$.

$\left( \frac{1}{3} \right)^{-2} = \left( \frac{3}{1} \right)^{2} = 3^2 = 9$.

$\left( \frac{1}{4} \right)^{-2} = \left( \frac{4}{1} \right)^{2} = 4^2 = 16$.

Now, add the results:

$4 + 9 + 16 = 13 + 16 = 29$.

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(iv) Evaluate $(3^{-1} + 4^{-1} + 5^{-1})^0$.

Use the rule $a^0 = 1$ for any non-zero base $a$.

The base of the exponent 0 is the expression inside the parentheses: $(3^{-1} + 4^{-1} + 5^{-1})$.

First, check if the base is non-zero.

$3^{-1} = \frac{1}{3}$, $4^{-1} = \frac{1}{4}$, $5^{-1} = \frac{1}{5}$.

The sum is $\frac{1}{3} + \frac{1}{4} + \frac{1}{5}$. This sum is clearly not zero.

Therefore, any non-zero base raised to the power of 0 is 1.

$(3^{-1} + 4^{-1} + 5^{-1})^0 = 1$.

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(v) Evaluate $\left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^{2}$.

Use the power of a power rule $(a^m)^n = a^{mn}$ with $a = \left( \frac{-2}{3} \right)^{-2}$, $m=1$, and $n=2$. This gives $\left( \left( \frac{-2}{3} \right)^{-2} \right)^2 = \left( \frac{-2}{3} \right)^{-2 \times 2} = \left( \frac{-2}{3} \right)^{-4}$.

Alternatively, first evaluate the inner expression using the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$:

$\left( \frac{-2}{3} \right)^{-2} = \left( \frac{3}{-2} \right)^{2}$.

The expression becomes $\left\{ \left( \frac{3}{-2} \right)^{2} \right\}^{2}$.

Now, use the power of a power rule $(a^m)^n = a^{mn}$ with $a = \frac{3}{-2}$, $m=2$, and $n=2$.

$\left( \left( \frac{3}{-2} \right)^{2} \right)^{2} = \left( \frac{3}{-2} \right)^{2 \times 2} = \left( \frac{3}{-2} \right)^{4}$.

Apply the power to the numerator and the denominator:

$\left( \frac{3}{-2} \right)^{4} = \frac{3^4}{(-2)^4}$.

Calculate the values:

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

$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2) = 16$ (Since the base is negative and the exponent is even).

So, $\frac{3^4}{(-2)^4} = \frac{81}{16}$.

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Conclusion:

(i) $(3^0 + 4^{-1}) \times 2^2 = 5$.

(ii) $(2^{-1} \times 4^{-1}) \div 2^{-2} = \frac{1}{2}$.

(iii) $\left( \frac{1}{2} \right)^{-2}$ + $\left( \frac{1}{3} \right)^{-2}$ + $\left( \frac{1}{4} \right)^{-2} = 29$.

(iv) $(3^{-1} + 4^{-1} + 5^{-1})^0 = 1$.

(v) $\left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^{2} = \frac{81}{16}$.

Given:

The expressions to be evaluated.

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To Evaluate:

The value of each expression.

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Solution:

(i) Evaluate $\frac{8^{-1} \;\times\; 5^{3}}{2^{-4}}$.

Use the rule $a^{-m} = \frac{1}{a^m}$ and $\frac{1}{a^{-m}} = a^m$:

$\frac{8^{-1} \;\times\; 5^{3}}{2^{-4}} = 8^{-1} \times 5^3 \times 2^{-(-4)}$

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

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

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

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

$= 2^{-3} \times 5^3 \times 2^4$

Group terms with the same base and use the rule $a^m \times a^n = a^{m+n}$:

$= (2^{-3} \times 2^4) \times 5^3$

$= 2^{-3+4} \times 5^3$

$= 2^1 \times 5^3$

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

$= 2 \times 125$

$= 250$.

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(ii) Evaluate $(5^{-1} \times 2^{-1}) \times 6^{-1}$.

Use the rule $a^{-m} = \frac{1}{a^m}$ for each term:

$(5^{-1} \times 2^{-1}) \times 6^{-1} = \left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}$

Multiply the terms inside the parentheses:

$= \left(\frac{1 \times 1}{5 \times 2}\right) \times \frac{1}{6}$

$= \frac{1}{10} \times \frac{1}{6}$

Multiply the resulting fractions:

$= \frac{1 \times 1}{10 \times 6}$

$= \frac{1}{60}$.

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Alternate Method for (ii):

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

$(5^{-1} \times 2^{-1}) = (5 \times 2)^{-1} = 10^{-1}$.

The expression becomes $10^{-1} \times 6^{-1}$.

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

$10^{-1} \times 6^{-1} = (10 \times 6)^{-1} = 60^{-1}$.

Use the rule $a^{-m} = \frac{1}{a^m}$:

$60^{-1} = \frac{1}{60^1} = \frac{1}{60}$.

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Conclusion:

(i) The value of $\frac{8^{-1} \;\times\; 5^{3}}{2^{-4}}$ is $250$.

(ii) The value of $(5^{-1} \times 2^{-1}) \times 6^{-1}$ is $\frac{1}{60}$.

Given:

The equation $5^{m} \div 5^{-3} = 5^{5}$.

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To Find:

The value of $m$.

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Solution:

We are given the equation:

The left side of the equation involves the division of two terms with the same base (5). We can use the quotient rule of exponents, which states that $a^m \div a^n = a^{m-n}$.

Applying the quotient rule to the left side of equation (1):

5$^{m}$ ÷ 5$^{-3} = 5^{m - (-3)}$

5$^{m}$ ÷ 5$^{-3} = 5^{m + 3}$.

Now, substitute this back into the original equation (1):

Equation (2) states that two exponential expressions with the same non-zero base are equal. This implies that their exponents must be equal.

Therefore, we can equate the exponents:

$m + 3 = 5$

Solve for $m$ by subtracting 3 from both sides of the equation:

$m = 5 - 3$

$m = 2$.

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Conclusion:

The value of $m$ is $\boxed{2}$.

Given:

The expressions to be evaluated.

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To Evaluate:

The value of each expression.

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Solution:

(i) Evaluate $\left\{ \left( \frac{1}{3} \right)^{-1}-\left( \frac{1}{4} \right)^{-1} \right\}^{-1}$.

First, evaluate the terms inside the curly braces with negative exponents using the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$:

$\left( \frac{1}{3} \right)^{-1} = \left( \frac{3}{1} \right)^{1} = 3^1 = 3$.

$\left( \frac{1}{4} \right)^{-1} = \left( \frac{4}{1} \right)^{1} = 4^1 = 4$.

The expression inside the curly braces is now $3 - 4 = -1$.

The entire expression is now $(-1)^{-1}$.

Using the rule $a^{-m} = \frac{1}{a^m}$:

$(-1)^{-1} = \frac{1}{(-1)^1} = \frac{1}{-1} = -1$.

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(ii) Evaluate $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-4}$.

We have two terms multiplied together. Notice that the bases $\frac{5}{8}$ and $\frac{8}{5}$ are reciprocals of each other.

Use the rule $\left( \frac{a}{b} \right)^{-m} = \left( \frac{b}{a} \right)^m$ on the first term:

$\left( \frac{5}{8} \right)^{-7} = \left( \frac{8}{5} \right)^{7}$.

Now the expression becomes $\left( \frac{8}{5} \right)^{7} \times \left( \frac{8}{5} \right)^{-4}$.

We have the multiplication of two terms with the same base ($\frac{8}{5}$). Use the product rule of exponents ($a^m \times a^n = a^{m+n}$):

$\left( \frac{8}{5} \right)^{7} \times \left( \frac{8}{5} \right)^{-4} = \left( \frac{8}{5} \right)^{7 + (–4)}$

$\left( \frac{8}{5} \right)^{7 + (–4)} = \left( \frac{8}{5} \right)^{7 - 4} = \left( \frac{8}{5} \right)^{3}$.

Now, apply the power to the numerator and the denominator:

$\left( \frac{8}{5} \right)^{3} = \frac{8^3}{5^3}$

Calculate the values:

$8^3 = 8 \times 8 \times 8 = 64 \times 8 = 512$.

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

So, $\frac{8^3}{5^3} = \frac{512}{125}$.

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Conclusion:

(i) The value of $\left\{ \left( \frac{1}{3} \right)^{-1}-\left( \frac{1}{4} \right)^{-1} \right\}^{-1}$ is $-1$.

(ii) The value of $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-4}$ is $\frac{512}{125}$.

Given:

The expressions to be simplified.

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To Simplify:

Simplify the given expressions.

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Solution:

(i) Simplify $\frac{25 \;\times\; t^{-4}}{5^{-3} \;\times\; 10 \;\times\; t^{-8}}$ (t ≠ 0).

Express the numbers in terms of their prime factors:

$25 = 5^2$

$10 = 2 \times 5$

Substitute these into the expression:

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

Rearrange the terms and group by base:

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

Combine the terms with the same base in the denominator using the product rule ($a^m \times a^n = a^{m+n}$):

Denominator: $2^1 \times 5^{-3+1} \times t^{-8} = 2^1 \times 5^{-2} \times t^{-8}$.

The expression is now:

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

Use the quotient rule ($a^m \div a^n = a^{m-n}$) for each base:

For base 5: $\frac{5^2}{5^{-2}} = 5^{2 - (-2)} = 5^{2+2} = 5^4$.

For base $t$: $\frac{t^{-4}}{t^{-8}} = t^{-4 - (-8)} = t^{-4 + 8} = t^4$.

For base 2: $\frac{1}{2^1} = 2^{-1}$.

Combine the simplified terms:

$5^4 \times t^4 \times 2^{-1}$

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

$= \frac{625 \times t^4}{2}$.

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(ii) Simplify $\frac{3^{-5} \;\times\; 10^{-5} \;\times\; 125}{5^{-7} \;\times\; 6^{-5}}$.

Express the numbers in terms of their prime factors:

$10 = 2 \times 5$

$125 = 5^3$

$6 = 2 \times 3$

Substitute these into the expression:

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

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

$(2 \times 5)^{-5} = 2^{-5} \times 5^{-5}$.

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

Substitute these back into the expression:

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

Rearrange the terms and group by base:

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

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

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

The expression is now:

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

Use the quotient rule ($a^m \div a^n = a^{m-n}$) for each base:

For base 3: $\frac{3^{-5}}{3^{-5}} = 3^{-5 - (-5)} = 3^{-5 + 5} = 3^0 = 1$.

For base 2: $\frac{2^{-5}}{2^{-5}} = 2^{-5 - (-5)} = 2^{-5 + 5} = 2^0 = 1$.

For base 5: $\frac{5^{-2}}{5^{-7}} = 5^{-2 - (-7)} = 5^{-2 + 7} = 5^5$.

Combine the simplified terms:

$1 \times 1 \times 5^5 = 5^5$.

Calculate the value of $5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 25 \times 25 \times 5 \ $$ = 625 \times 5 = 3125$.

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Conclusion:

(i) $\frac{25 \;\times\; t^{-4}}{5^{-3} \;\times\; 10 \;\times\; t^{-8}} = \frac{625 t^4}{2}$.

(ii) $\frac{3^{-5} \;\times\; 10^{-5} \;\times\; 125}{5^{-7} \;\times\; 6^{-5}} = 3125$.

Concept Used:

Standard form (or scientific notation) of a number is expressed as a number between 1.0 and 10.0 multiplied by a power of 10.

A number $N$ in standard form is written as $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

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Solution:

(i) Express 0.000035 in standard form.

We need to move the decimal point to the right until it is after the first non-zero digit (which is 3).

0.000035

Move the decimal point 5 places to the right: $3.5$.

Since we moved the decimal point to the right, the power of 10 will be negative. The number of places moved is 5, so the power is $-5$.

So, $0.000035 = 3.5 \times 10^{-5}$.

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(ii) Express 4050000 in standard form.

We need to move the decimal point to the left until it is after the first non-zero digit (which is 4).

4050000. (The decimal point is implicitly at the end of the integer).

Move the decimal point 6 places to the left: $4.05$.

Since we moved the decimal point to the left, the power of 10 will be positive. The number of places moved is 6, so the power is $6$.

So, $4050000 = 4.05 \times 10^{6}$.

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Conclusion:

(i) The standard form of 0.000035 is $3.5 \times 10^{-5}$.

(ii) The standard form of 4050000 is $4.05 \times 10^{6}$.

To convert a number from scientific notation ($a \times 10^n$) to usual form:

If $n$ is positive, move the decimal point $n$ places to the right.

If $n$ is negative, move the decimal point $|n|$ places to the left.

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(i) We are given the number $3.52 \times 10^5$.

Here, the power of 10 is $5$, which is positive.

So, we move the decimal point in $3.52$ five places to the right.

$3.52 \times 10^5 = 3.52000 \times 10^5$

Moving the decimal point 5 places to the right gives:

$352000$.

The usual form of $3.52 \times 10^5$ is $352000$.

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(ii) We are given the number $7.54 \times 10^{-4}$.

Here, the power of 10 is $-4$, which is negative.

So, we move the decimal point in $7.54$ four places to the left.

$7.54 \times 10^{-4} = 00007.54 \times 10^{-4}$

Moving the decimal point 4 places to the left gives:

$0.000754$.

The usual form of $7.54 \times 10^{-4}$ is $0.000754$.

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(iii) We are given the number $3 \times 10^{-5}$.

Here, the power of 10 is $-5$, which is negative.

So, we move the decimal point in $3$ (which is $3.0$) five places to the left.

$3 \times 10^{-5} = 000003.0 \times 10^{-5}$

Moving the decimal point 5 places to the left gives:

$0.00003$.

The usual form of $3 \times 10^{-5}$ is $0.00003$.

To express a number in standard form ($a \times 10^n$), where $1 \leq a < 10$ and $n$ is an integer:

Move the decimal point so that there is only one non-zero digit to the left of the decimal point.

The number of places the decimal point is moved is the exponent $n$.

If the decimal point is moved to the left, $n$ is positive. If it is moved to the right, $n$ is negative.

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(i) We need to express $0.0000000000085$ in standard form.

Move the decimal point 12 places to the right to get $8.5$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 12.

So, $n = -12$.

The standard form is $8.5 \times 10^{-12}$.

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(ii) We need to express $0.00000000000942$ in standard form.

Move the decimal point 12 places to the right to get $9.42$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 12.

So, $n = -12$.

The standard form is $9.42 \times 10^{-12}$.

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(iii) We need to express $6020000000000000$ in standard form.

Assume the decimal point is at the end of the number ($6020000000000000.$).

Move the decimal point 15 places to the left to get $6.02$.

Since the decimal point was moved to the left, the exponent is positive. The number of places moved is 15.

So, $n = 15$.

The standard form is $6.02 \times 10^{15}$.

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(iv) We need to express $0.00000000837$ in standard form.

Move the decimal point 9 places to the right to get $8.37$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 9.

So, $n = -9$.

The standard form is $8.37 \times 10^{-9}$.

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(v) We need to express $31860000000$ in standard form.

Assume the decimal point is at the end of the number ($31860000000.$).

Move the decimal point 10 places to the left to get $3.186$.

Since the decimal point was moved to the left, the exponent is positive. The number of places moved is 10.

So, $n = 10$.

The standard form is $3.186 \times 10^{10}$.

To convert a number from standard form ($a \times 10^n$) to usual form:

If $n$ is positive, move the decimal point $n$ places to the right.

If $n$ is negative, move the decimal point $|n|$ places to the left.

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(i) We are given the number $3.02 \times 10^{-6}$.

Here, the power of 10 is $-6$, which is negative.

So, we move the decimal point in $3.02$ six places to the left.

$3.02 \times 10^{-6} = 0000003.02 \times 10^{-6}$

Moving the decimal point 6 places to the left gives:

$0.00000302$.

The usual form of $3.02 \times 10^{-6}$ is $0.00000302$.

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(ii) We are given the number $4.5 \times 10^{4}$.

Here, the power of 10 is $4$, which is positive.

So, we move the decimal point in $4.5$ four places to the right.

$4.5 \times 10^{4} = 4.5000 \times 10^{4}$

Moving the decimal point 4 places to the right gives:

$45000$.

The usual form of $4.5 \times 10^{4}$ is $45000$.

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(iii) We are given the number $3 \times 10^{-8}$.

Here, the power of 10 is $-8$, which is negative.

So, we move the decimal point in $3$ (which is $3.0$) eight places to the left.

$3 \times 10^{-8} = 000000003.0 \times 10^{-8}$

Moving the decimal point 8 places to the left gives:

$0.00000003$.

The usual form of $3 \times 10^{-8}$ is $0.00000003$.

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(iv) We are given the number $1.0001 \times 10^{9}$.

Here, the power of 10 is $9$, which is positive.

So, we move the decimal point in $1.0001$ nine places to the right.

$1.0001 \times 10^{9} = 1.000100000 \times 10^{9}$

Moving the decimal point 9 places to the right gives:

$1000100000$.

The usual form of $1.0001 \times 10^{9}$ is $1000100000$.

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(v) We are given the number $5.8 \times 10^{12}$.

Here, the power of 10 is $12$, which is positive.

So, we move the decimal point in $5.8$ twelve places to the right.

$5.8 \times 10^{12} = 5.800000000000 \times 10^{12}$

Moving the decimal point 12 places to the right gives:

$5800000000000$.

The usual form of $5.8 \times 10^{12}$ is $5800000000000$.

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(vi) We are given the number $3.61492 \times 10^{6}$.

Here, the power of 10 is $6$, which is positive.

So, we move the decimal point in $3.61492$ six places to the right.

$3.61492 \times 10^{6} = 3.614920 \times 10^{6}$

Moving the decimal point 6 places to the right gives:

$3614920$.

The usual form of $3.61492 \times 10^{6}$ is $3614920$.

To express a number in standard form ($a \times 10^n$), where $1 \leq a < 10$ and $n$ is an integer, we move the decimal point so that there is only one non-zero digit to the left of the decimal point. The exponent $n$ is the number of places the decimal point was moved; it is positive if moved to the left and negative if moved to the right.

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(i) The number is $\frac{1}{1000000}$ m.

We can write this as $1 \div 1000000$.

$\frac{1}{1000000} = \frac{1}{10^6}$

Using the law of exponents $a^{-m} = \frac{1}{a^m}$, we get:

$\frac{1}{10^6} = 1 \times 10^{-6}$.

The number $1$ is already between $1$ and $10$.

The standard form of $\frac{1}{1000000}$ m is $1 \times 10^{-6}$ m.

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(ii) The number representing the charge of an electron is $0.00000000000000000016$ coulomb.

Move the decimal point 19 places to the right to get $1.6$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 19.

So, $n = -19$.

The standard form is $1.6 \times 10^{-19}$ coulomb.

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(iii) The number representing the size of a bacteria is $0.0000005$ m.

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

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 7.

So, $n = -7$.

The standard form is $5 \times 10^{-7}$ m.

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(iv) The number representing the size of a plant cell is $0.00001275$ m.

Move the decimal point 5 places to the right to get $1.275$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 5.

So, $n = -5$.

The standard form is $1.275 \times 10^{-5}$ m.

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(v) The number representing the thickness of a thick paper is $0.07$ mm.

Move the decimal point 2 places to the right to get $7$.

Since the decimal point was moved to the right, the exponent is negative. The number of places moved is 2.

So, $n = -2$.

The standard form is $7 \times 10^{-2}$ mm.

Given:

Number of books in the stack = 5

Thickness of each book = 20 mm

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Total thickness of 5 books = Number of books $\times$ Thickness of each book

Total thickness of books = $5 \times 20$ mm

Total thickness of books = $100$ mm

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Given:

Number of paper sheets in the stack = 5

Thickness of each paper sheet = 0.016 mm

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Total thickness of 5 paper sheets = Number of paper sheets $\times$ Thickness of each paper sheet

Total thickness of paper sheets = $5 \times 0.016$ mm

Total thickness of paper sheets = $0.080$ mm

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Total thickness of the stack = Total thickness of books + Total thickness of paper sheets

Total thickness of the stack = $100$ mm + $0.080$ mm

Total thickness of the stack = $100.08$ mm

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To express the total thickness in standard form:

The total thickness is $100.08$ mm.

Move the decimal point 2 places to the left to get $1.0008$.

Since the decimal point was moved 2 places to the left, the power of 10 is $2$.

The standard form is $1.0008 \times 10^2$.

Therefore, the total thickness of the stack is $100.08$ mm or $1.0008 \times 10^2$ mm in standard form.