# Powers - Class 8 - Mathematics

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## Examples - Powers | R.D. Sharma | Mathematics | Class 8

Express each of the following as a rational number of the form $\frac{p}{q}$ :

(i) $5^{-3}$

(ii) $(-2)^{-5}$

(iii) $\left(\frac{4}{3}\right)^{-3}$

(iv) $\left(\frac{-2}{5}\right)^{-4}$

(v) $\frac{1}{2^{-3}}$

To express each of the given expressions as a rational number of the form $\frac{p}{q}$, we use the property that $a^{-n} = \frac{1}{a^n}$ for any nonzero $a$ and positive integer $n$. Therefore:

(i) $$5^{-3} = \frac{1}{5^3} = \frac{1}{125}$$

(ii) $$(-2)^{-5} = \frac{1}{(-2)^5} = \frac{1}{-32}$$

(iii) $$\left(\frac{4}{3}\right)^{-3} = \frac{1}{\left(\frac{4}{3}\right)^3} = \frac{1}{\frac{64}{27}} = \frac{27}{64}$$

(iv) $$\left(\frac{-2}{5}\right)^{-4} = \frac{1}{\left(\frac{-2}{5}\right)^4} = \frac{1}{\frac{16}{625}} = \frac{625}{16}$$

(v) $$\frac{1}{2^{-3}} = 2^3 = 8$$

Hence, the rational expressions are:

(i) $\frac{1}{125}$

(ii) $\frac{1}{-32}$

(iii) $\frac{27}{64}$

(iv) $\frac{625}{16}$

(v) $8$, which can also be written as $\frac{8}{1}$ for consistency with the $\frac{p}{q}$ form.

Express each of the following as a rational number of the form $\frac{p}{q}$ :

(i) $\left(\frac{3}{8}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}$

(ii) $\left(\frac{-2}{7}\right)^{-4} \times\left(\frac{-7}{5}\right)^{2}$

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Sign up nowExpress each of the following as power of a rational number with positiv exponent:

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

(ii) $5^{-3} \times 5^{-6}$

(iii) $\left(\frac{-1}{4}\right)^{-5} \times\left(\frac{-1}{4}\right)^{-7}$

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Sign up nowSimplify, with steps

(i) $\left(2^{-1}+5^{-1}\right)^{2} \times\left(\frac{-5}{8}\right)^{-1}$

(ii) $\left(6^{-1}-8^{-1}\right)^{-1}+\left(2^{-1}-3^{-1}\right)^{-1}$

(iii) $\left(5^{-1} \times 3^{-1}\right)^{-1}+6^{-1}$

(iv) $\left(4^{-1}+8^{-1}\right)+\left(\frac{2}{3}\right)^{-1}$

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Sign up nowSimplify, with steps

(i) $(\frac{1}{4})^{-2} + (\frac{1}{2})^{-2} + (\frac{1}{3})^{-2}$

(ii) $\left\{6^{-1} + (\frac{3}{2})^{-1}\right\}^{-1}$

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Express each of the following as a rational number of the form $\frac{p}{q}$ :

(i) $\left(2^{-1}+3^{-1}\right)^{2}$

(ii) $\left(2^{-1}-4^{-1}\right)^{2}$

(iii) $\left\{\left(\frac{3}{4}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$

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By what number should $(-8)^{-1}$ be multiplied so that the product may be equal to $10^{-1}$ ?

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By what number should $(-24)^{-1}$ be divided so that the quotient may be $3^{-1}$ ?

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Using the laws of exponents, simplify each of the following and express in exponential form:

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

(ii) $2^{-7} \div 2^{-3}$

(iii) $\left(5^{2}\right)^{-3}$

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

(v) $\frac{3^{-5}}{4^{-5}}$

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Using the laws of exponents simplify and express each of the following in exponential form with positive exponent, with steps

(i) $(-4)^{4} \times(-4)^{-10}$

(ii) $2^{-5}+2^{2}$

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

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

(v) $\left(3^{-7}+3^{-10}\right) \times 3^{-5}$

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

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Simplify and write the answer in the exponential form, with steps

(i) $\left(2^{5} \div 2^{8}\right)^{5} \times 2^{-5}$

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

(iii) $\frac{1}{8} \times 3^{-3}$

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Simplify each of the following:

(i) $\left[\left\{\left(\frac{-1}{5}\right)^{-2}\right\}^{2}\right]^{-1}$

(ii) $\left\{\left(\frac{1}{3}\right)^{-2}-\left(\frac{1}{2}\right)^{-3}\right\} \div\left(\frac{1}{4}\right)^{-2}$

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Simplify:

(i) $\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-5}$

(ii) $\left(\frac{-2}{3}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}$

(iii) $\left(\frac{3}{4}\right)^{-4} \div\left(\frac{3}{2}\right)^{-3}$

(iv) $\left(\frac{3}{7}\right)^{-2} \times\left(\frac{7}{6}\right)^{-3}$

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Evaluate: $\frac{8^{-1} \times 5^{3}}{2^{-4}}$, with steps

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Simplify, with steps

(i) $\frac{25 \times a^{-4}}{5^{-3} \times 10 \times a^{-8}}$

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

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By what number should $(-4)^{-2}$ be multiplied so that the product may be equal to $10^{-2}$?

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By what number should $(-12)^{-1}$ be divided so that the quotient may be $\left(\frac{2}{3}\right)^{-1}$ ?

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By what number should $\left(\frac{-3}{2}\right)^{-3}$ be divided so that the quotient may be

$$ \left(\frac{4}{27}\right)^{-2} ? $$

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Find $x$ so that $\left(\frac{5}{3}\right)^{-5} \times\left(\frac{5}{3}\right)^{-11}=\left(\frac{5}{3}\right)^{8 x}$

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Find $m$ so that $\left(\frac{2}{9}\right)^{3} \times\left(\frac{2}{9}\right)^{-6}=\left(\frac{2}{9}\right)^{2 m-1}$

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If $x=\left(\frac{3}{2}\right)^{2} \times\left(\frac{2}{3}\right)^{-4}$, find the value of $x^{-2}$, with steps

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Write the following numbers in standard form:

(i) 0.4579

(ii) 0.000007

(iii) 0.000000564

(iv) 0.0000021

(v) 216000000

(vi) $0.0000529 \times 10^{4}$

(vii) $9573 \times 10^{-4}$

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Express the following numbers in usual form:

(i) $3.52 \times 10^{5}$

(ii) $7.54 \times 10^{-4}$

(iii) $3 \times 10^{-5}$

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Express the number appearing in the following statements in standard form:

(i) 1 micron is equal to $\frac{1}{1000000}$ metre.

(ii) Charge of an electron is 0.0000000000000000016 coloumbs.

(iii) Size of a bacteria is 0.0000005 metre.

(iv) Size of a plant cell is 0.00001275 metre.

(v) Thickness of a normal paper is $0.07 \mathrm{~mm}$.

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If the diameters of the Sun and the Earth are $1.4 \times 10^{9}$ metres and $1.275 \times 10^{7}$ metres respectively. Compare these two.

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The size of a red blood cell is $0.000007 \mathrm{~m}$ and the size of a plant cell is $0.00001275 \mathrm{~m}$. Compare these two.

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## Exercise 2.1 - Powers | R.D. Sharma | Mathematics | Class 8

Express each of the following as a rational number of the form $\frac{p}{q}$, where $p$ and $q$ ar integers and $q \neq 0$ :

(i) $2^{-3}$

(ii) $(-4)^{-2}$

(iii) $\frac{1}{3^{-2}}$

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

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

Here are the expressions expressed as rational numbers of the form $\frac{p}{q}$:

For $2^{-3}$, it is expressed as $\frac{1}{8}$.

For $(-4)^{-2}$, it is expressed as $\frac{1}{16}$.

For $\frac{1}{3^{-2}}$, it is expressed as $9$ or $\frac{9}{1}$.

For $\left(\frac{1}{2}\right)^{-5}$, it is expressed as $32$ or $\frac{32}{1}$.

For $\left(\frac{2}{3}\right)^{-2}$, it is expressed as $\frac{9}{4}$.

Find the values of each of the following, with steps:

(i) $3^{-1}+4^{-1}$

(ii) $\left(3^{0}+4^{-1}\right) \times 2^{2}$

(iii) $\left(3^{-1}+4^{-1}+5^{-1}\right)^{0}$

(iv) $\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$

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Find the values of each of the following, with steps:

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

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

(iii) $\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}$

(iv) $\left(5^{-1} \times 2^{-1}\right) \div 6^{-1}$

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Simplify, with steps for each:

(i) $\left(4^{-1} \times 3^{-1}\right)^{2}$

(ii) $\left(5^{-1} \div 6^{-1}\right)^{3}$

(iii) $\left(2^{-1}+3^{-1}\right)^{-1}$

(iv) $\left(3^{-1} \times 4^{-1}\right)^{-1} \times 5^{-1}$

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Simplify, with steps for each:

(i) $\left(3^{2}+2^{2}\right) \times\left(\frac{1}{2}\right)^{3}$

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

(iii) $\left[\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right] \div\left(\frac{1}{4}\right)^{-3}$

(iv) $\left(2^{2}+3^{2}-4^{2}\right) \div\left(\frac{3}{2}\right)^{2}$

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By what number should $5^{-1}$ be multiplied so that the product may be equal to $(-7)^{-1}$ ?

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By what number should $\left(\frac{1}{2}\right)^{-1}$ be multiplied so that the product may be equal t $\left(-\frac{4}{7}\right)^{-1} ?$

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By what number should $(-15)^{-1}$ be divided so that the quotient may be equal to $(-5)^{-1}$.

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## Exercise 2.2 - Powers | R.D. Sharma | Mathematics | Class 8

Write each of the following in exponential form:

(i) $\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1}$

(ii) $\left(\frac{2}{5}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2}$

To write the given expressions in exponential form, we apply the rule of exponents that states if we multiply powers with the same base, we add the exponents. Let's simplify each expression:

(i) $\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1} \times\left(\frac{3}{2}\right)^{-1}$

Since we're multiplying the expression $\left(\frac{3}{2}\right)^{-1}$ four times, we add the exponents:

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

(ii) $\left(\frac{2}{5}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2}$

Similarly, we're multiplying the expression $\left(\frac{2}{5}\right)^{-2}$ three times, so we add the exponents:

The result is $\left(\frac{2}{5}\right)^{-6}$.

Evaluate, with steps

(i) $5^{-2}$

(ii) $(-3)^{-2}$

(iii) $\left(\frac{1}{3}\right)^{-4}$

(iv) $\left(\frac{-1}{2}\right)^{-1}$

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Express each of the following as a rational number in the form $\frac{p}{q}$ :

(i) $6^{-1}$

(ii) $(-7)^{-1}$

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

(iv) $(-4)^{-1} \times\left(\frac{-3}{2}\right)^{-1}$

(v) $\left(\frac{3}{5}\right)^{-1} \times\left(\frac{5}{2}\right)^{-1}$

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Simplify, with steps

(i) $\left\{4^{-1} \times 3^{-1}\right\}^{2}$

(ii) $\left\{5^{-1} \div 6^{-1}\right\}^{3}$

(iii) $\left(2^{-1}+3^{-1}\right)^{-1}$

(iv) $\left\{3^{-1} \times 4^{-1}\right\}^{-1} \times 5^{-1}$

(v) $\left(4^{-1}-5^{-1}\right) \div 3^{-1}$

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Express each of the following rational numbers with a negative exponent:

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

(ii) $3^{5}$

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

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

(v) $\left\{\left(\frac{7}{3}\right)^{4}\right\}^{-3}$

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Express each of the following rational numbers with a positive exponent, with steps

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

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

(iii) $4^{3} \times 4^{-9}$

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

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

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Simplify, with steps

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

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

(iii) $\left\{\left(\frac{1}{2}\right)^{-1} \times(-4)^{-1}\right\}^{-1}$

(iv) $\left[\left\{\left(\frac{-1}{4}\right)^{2}\right\}^{-2}\right]^{-1}$

(v) $\left\{\left(\frac{2}{3}\right)^{2}\right\}^{3} \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times 6^{-1}$

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By what number should $5^{-1}$ be multiplied so that the product may be equal to $(-7)^{-1}$ ?

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By what number should $\left(\frac{1}{2}\right)^{-1}$ be multiplied so that the product may be equal to $\left(\frac{-4}{7}\right)^{-1} ?$

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By what number should $(-15)^{-1}$ be divided so that the quotient may be equal to $(-5)^{-1}$ ?

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By what number should $\left(\frac{5}{3}\right)^{-2}$ be multiplied so that the product may be $\left(\frac{7}{3}\right)^{-1}$ ?

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Find $x$, with steps, if

(i) $\left(\frac{1}{4}\right)^{-4} \times\left(\frac{1}{4}\right)^{-8}=\left(\frac{1}{4}\right)^{-4 x}$

(ii) $\left(\frac{-1}{2}\right)^{-19} \div\left(\frac{-1}{2}\right)^{8}=\left(\frac{-1}{2}\right)^{-2 x+1}$

(iii) $\left(\frac{3}{2}\right)^{-3} \times\left(\frac{3}{2}\right)^{5}=\left(\frac{3}{2}\right)^{2 x+1}$

(iv) $\left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^{15}=\left(\frac{2}{5}\right)^{2+3 x}$

(v) $\left(\frac{5}{4}\right)^{-x} \div\left(\frac{5}{4}\right)^{-4}=\left(\frac{5}{4}\right)^{5}$

(vi) $\left(\frac{8}{3}\right)^{2 x+1} \times\left(\frac{8}{3}\right)^{5}=\left(\frac{8}{3}\right)^{x+2}$

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(i) If $x=\left(\frac{3}{2}\right)^{2} \times\left(\frac{2}{3}\right)^{-4}$, find the value of $x^{-2}$.

(ii) If $x=\left(\frac{4}{5}\right)^{-2} \div\left(\frac{1}{4}\right)^{2}$, find the value of $x^{-1}$.

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Find the value of $x$ for which $5^{2 x} \div 5^{-3}=5^{5}$.

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## Exercise 2.3 - Powers | R.D. Sharma | Mathematics | Class 8

Express the following numbers in standard form:

(i) 6020000000000000

(ii) 0.00000000000942

(iii) 0.00000000085

(iv) $846 \times 10^{7}$

(v) $3759 \times 10^{-4}$

(vi) 0.00072984

(vii) $0.000437 \times 10^{4}$

(viii) $4 \div 100000$

Here are the expressions in standard form:

(i) $6.02 \times 10^{15}$

(ii) $9.42 \times 10^{-12}$

(iii) $8.5 \times 10^{-10}$

(iv) $8.46 \times 10^{9}$

(v) $3.759 \times 10^{-1}$

(vi) $7.2984 \times 10^{-4}$

(vii) $4.37 \times 10^{0}$ (or simply 4.37)

(viii) $4.0 \times 10^{-5}$

Write the following numbers in the usual form:

(i) $4.83 \times 10^{7}$

(ii) $3.02 \times 10^{-6}$

(iii) $4.5 \times 10^{4}$

(iv) $3 \times 10^{-8}$

(v) $1.0001 \times 10^{9}$

(vi) $5.8 \times 10^{2}$

(vii) $3.61492 \times 10^{6}$

(viii) $3.25 \times 10^{-7}$

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## Exercise (MCQs) - Powers | R.D. Sharma | Mathematics | Class 8

Square of $\left(\frac{-2}{3}\right)$ is

(a) $-\frac{2}{3}$

(b) $\frac{2}{3}$

(c) $-\frac{4}{9}$

(d) $\frac{4}{9}$

To find the square of $\left(\frac{-2}{3}\right)$, we calculate $\left(\frac{-2}{3}\right)^2$.

$$ \left(\frac{-2}{3}\right)^2 = \frac{(-2)^2}{3^2} = \frac{4}{9} $$

Therefore, the correct answer is (d) $\frac{4}{9}$.

Cube of $\frac{-1}{2}$ is

(a) $\frac{1}{8}$

(b) $\frac{1}{16}$

(c) $-\frac{1}{8}$

(d) $\frac{-1}{16}$

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Which of the following is not equal to $\left(\frac{-3}{5}\right)^{4}$ ?

(a) $\frac{(-3)^{4}}{5^{4}}$

(b) $\frac{3^{4}}{(-5)^{4}}$

(c) $-\frac{3^{4}}{5^{4}}$

(d) $\frac{-3}{5} \times \frac{-3}{5} \times \frac{-3}{5} \times \frac{-3}{5}$

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Which of the following is not reciprocal of $\left(\frac{2}{3}\right)^{4}$ ?

(a) $\left(\frac{3}{2}\right)^{4}$

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

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

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

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Which of the following numbers is not equal to $\frac{-8}{27}$ ?

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

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

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

(d) $\left(\frac{-2}{3}\right) \times\left(\frac{-2}{3}\right) \times\left(\frac{-2}{3}\right)$

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$\left(\frac{2}{3}\right)^{-5}$ is equal to

(a) $\left(\frac{-2}{3}\right)^{5}$

(b) $\left(\frac{3}{2}\right)^{5}$

(c) $\frac{2 x-5}{3}$

(d) $\frac{2}{3 \times 5}$

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$\left(\frac{-1}{2}\right)^{5} \times\left(\frac{-1}{2}\right)^{3}$ is equal to

(a) $\left(\frac{-1}{2}\right)^{8}$

(b) $-\left(\frac{1}{2}\right)^{8}$

(c) $\left(\frac{1}{4}\right)^{8}$

(d) $\left(-\frac{1}{2}\right)^{15}$

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

(b) $\left(-\frac{1}{5}\right)^{11}$

(c) $(-5)^{5}$

(d) $\left(\frac{1}{5}\right)^{5}$

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$\left(\frac{-2}{5}\right)^{7} \div\left(\frac{-2}{5}\right)^{5}$ is equal to

(a) $\frac{4}{25}$

(b) $\frac{-4}{25}$

(c) $\left(\frac{-2}{5}\right)^{12}$

(d) $\frac{25}{4}$

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$\left\{\left(\frac{1}{3}\right)^{2}\right\}^{4}$ is equal to

(a) $\left(\frac{1}{3}\right)^{6}$

(b) $\left(\frac{1}{3}\right)^{8}$

(c) $\left(\frac{1}{3}\right)^{24}$

(d) $\left(\frac{1}{3}\right)^{16}$

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$\left(\frac{1}{5}\right)^{0}$ is equal to

(a) 0

(b) $\frac{1}{5}$

(c) 1

(d) 5

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$\left(\frac{-3}{2}\right)^{-1}$ is equal to

(a) $\frac{2}{3}$

(b) $-\frac{2}{3}$

(c) $\frac{3}{2}$

(d) none of these

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$\left(\frac{2}{3}\right)^{-5} \times\left(\frac{5}{7}\right)^{-5}$ is equal to

(a) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-10}$

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

(c) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{25}$

(d) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-25}$

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$\left(\frac{3}{4}\right)^{5} \div\left(\frac{5}{3}\right)^{5}$ is equal to

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

(b) $\left(\frac{3}{4} \div \frac{5}{3}\right)^{1}$

(c) $\left(\frac{3}{4} \div \frac{5}{3}\right)^{0}$

(d) $\left(\frac{3}{4} \div \frac{5}{3}\right)^{10}$

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For any two non-zero rational numbers $a$ and $b, a^{4} \div b^{4}$ is equal to

(a) $(a \div b)^{1}$

(b) $(a \div b)^{0}$

(c) $(a \div b)^{4}$

(d) $(a \div b)^{8}$

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For any two rational numbers $a$ and $b, a^{5} \times b^{5}$ is equal to

(a) $(a \times b)^{0}$

(b) $(a \times b)^{10}$

(c) $(a \times b)^{5}$

(d) $(a \times b)^{25}$

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For a non-zero rational number $a, a^{7} \div a^{12}$ is equal to

(a) $a^{5}$

(b) $a^{-19}$

(c) $a^{-5}$

(d) $a^{19}$

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For a non zero rational number $a,\left(a^{3}\right)^{-2}$ is equal to

(a) $a^{6}$

(b) $a^{-6}$

(c) $a^{-9}$

(d) $a^{1}$