Exponents and Powers — Exercise 2

Evaluate:

$\Big(\dfrac{3}{5}\Big)^{-2}$

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow \Big(\dfrac{3}{5}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{2}\\[1em] \Rightarrow \Big(\dfrac{5 \times 5}{3 \times 3}\Big)\\[1em] \Rightarrow \dfrac{25}{9}$

Hence, $\mathbf{\Big(\dfrac{3}{5}\Big)^{-2} = \dfrac{25}{9}}$.

Evaluate:

(-3)^-3

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow (-3)^{-3}\\[1em] \Rightarrow \dfrac{1}{(-3)^3}\\[1em] \Rightarrow \dfrac{1}{(-3) \times (-3) \times (-3)}\\[1em] \Rightarrow \dfrac{1}{-27}\\[1em] \Rightarrow -\dfrac{1}{27}$

Hence, $\mathbf{(-3)^{-3} = -\dfrac{1}{27}}$.

Evaluate:

$\Big(\dfrac{2}{7}\Big)^{-4}$

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow \Big(\dfrac{2}{7}\Big)^{-4}\\[1em] \Rightarrow \Big(\dfrac{7}{2}\Big)^{4}\\[1em] \Rightarrow \Big(\dfrac{7 \times 7 \times 7 \times 7}{2 \times 2 \times 2 \times 2}\Big)\\[1em] \Rightarrow \dfrac{2401}{16}$

Hence, $\mathbf{\Big(\dfrac{2}{7}\Big)^{-4} = \dfrac{2401}{16}}$.

Simplify:

[(2)^-1 + (4)^-1 + (3)^-1]^-1

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow [(2)^{-1} + (4)^{-1} + (3)^{-1}]^{-1}\\[1em] \Rightarrow \Big[\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{3}\Big]^{-1}$

LCM of 2, 4 and 3 is 2 x 2 x 3 = 12

$\Rightarrow \Big[\dfrac{1 \times 6}{2 \times 6} + \dfrac{1 \times 3}{4 \times 3} + \dfrac{1 \times 4}{3 \times 4}\Big]^{-1}\\[1em] \Rightarrow \Big[\dfrac{6}{12} + \dfrac{3}{12} + \dfrac{4}{12}\Big]^{-1}\\[1em] \Rightarrow \Big[\dfrac{6 + 3 + 4}{12}\Big]^{-1}\\[1em] \Rightarrow \Big[\dfrac{13}{12}\Big]^{-1}\\[1em] \Rightarrow \dfrac{12}{13}$

Hence, $\mathbf{[(2)^{-1} + (4)^{-1} + (3)^{-1}]^{-1} = \dfrac{12}{13}}$.

Simplify:

$[(4)^{-1} - (5)^{-1}]^2 \times \Big(\dfrac{5}{8}\Big)^{-1}$

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow [(4)^{-1} - (5)^{-1}]^2 \times \Big(\dfrac{5}{8}\Big)^{-1}\\[1em] \Rightarrow \Big[\dfrac{1}{4} - \dfrac{1}{5}\Big]^2 \times \Big(\dfrac{8}{5}\Big)$

LCM of 4 and 5 is 4 x 5 = 20

$\Rightarrow \Big[\dfrac{1 \times 5}{4 \times 5} - \dfrac{1 \times 4}{5 \times 4}\Big]^2 \times \Big(\dfrac{8}{5}\Big)\\[1em] \Rightarrow \Big[\dfrac{5}{20} - \dfrac{4}{20}\Big]^2 \times \Big(\dfrac{8}{5}\Big)\\[1em] \Rightarrow \Big[\dfrac{5 - 4}{20}\Big]^2 \times \Big(\dfrac{8}{5}\Big)\\[1em] \Rightarrow \Big[\dfrac{1}{20}\Big]^2 \times \Big(\dfrac{8}{5}\Big)\\[1em] \Rightarrow \dfrac{1 \times 1 \times 8}{20 \times 20 \times 5}\\[1em] \Rightarrow \dfrac{8}{2000}\\[1em] \Rightarrow \dfrac{1}{250}$

Hence, $\mathbf{[(4)^{-1}-(5)^{-1}]^2 \times \Big(\dfrac{5}{8}\Big)^{-1}=\dfrac{1}{250}}$.

Simplify:

[4⁰ + 4² - 2³] × 3^-2

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^0 = 1$.

Solving,

$\Rightarrow [4^0 + 4^2 - 2^3] \times 3^{-2}\\[1em] \Rightarrow [1 + (4 \times 4) - (2 \times 2 \times 2)] \times \dfrac{1}{3^2}\\[1em] \Rightarrow [1 + 16 - 8] \times \dfrac{1}{9}\\[1em] \Rightarrow [9] \times \dfrac{1}{9}\\[1em] \Rightarrow \dfrac{9}{9}\\[1em] \Rightarrow 1$

Hence, $\mathbf{[4^0 + 4^2 - 2^3] \times 3^{-2} = 1}$.

Simplify:

$\Big[(5)^2 - \Big(\dfrac{1}{4}\Big)^{-2}\Big] \times \Big(\dfrac{3}{4}\Big)^{-2}$

Answer

As we know, for any non-zero rational number $a$.

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

Solving,

$\Rightarrow \Big[(5)^2 - \Big(\dfrac{1}{4}\Big)^{-2}\Big] \times \Big(\dfrac{3}{4}\Big)^{-2}\\[1em] \Rightarrow \Big[(5 \times 5) - \Big(\dfrac{4}{1}\Big)^{2}\Big] \times \Big(\dfrac{4}{3}\Big)^{2}\\[1em] \Rightarrow \Big[25 - (4 \times 4)\Big] \times \Big(\dfrac{4 \times 4}{3 \times 3}\Big)\\[1em] \Rightarrow \Big[25 - 16\Big] \times \Big(\dfrac{16}{9}\Big)\\[1em] \Rightarrow [9] \times \Big(\dfrac{16}{9}\Big)\\[1em] \Rightarrow \dfrac{9 \times 16}{9}\\[1em] \Rightarrow 16$

Hence, $\mathbf{\Big[(5)^2 - \Big(\dfrac{1}{4}\Big)^{-2}\Big] \times \Big(\dfrac{3}{4}\Big)^{-2} = 16}$.

Find the multiplicative inverse of:

$\Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{81}{16}\Big)^{\dfrac{-3}{4}}\\[1em] \Rightarrow \Big(\dfrac{3^4}{2^4}\Big)^{\dfrac{-3}{4}}\\[1em] \Rightarrow \Big(\dfrac{3}{2}\Big)^{4 \times \dfrac{-3}{4}}\\[1em] \Rightarrow \Big(\dfrac{3}{2}\Big)^{-3}\\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{3}\\[1em] \Rightarrow \dfrac{2 \times 2 \times 2}{3 \times 3 \times 3}\\[1em] \Rightarrow \dfrac{8}{27}$

The multiplicative inverse of a non-zero number is its reciprocal.

Thus, multiplicative inverse of $\dfrac{8}{27} = \dfrac{27}{8}$.

Hence, the multiplicative inverse of $\mathbf{\Big(\dfrac{81}{16}\Big)^{\dfrac{-3}{4}}}$ is $\mathbf{\dfrac{27}{8}}$.

Find the multiplicative inverse of:

$\Big\lbrace\Big(\dfrac{-3}{2}\Big)^{-4}\Big\rbrace^{\dfrac{1}{2}}$

Answer

Solving,

$\Rightarrow \Big\lbrace\Big(\dfrac{-3}{2}\Big)^{-4}\Big\rbrace^{\dfrac{1}{2}}\\[1em] \Rightarrow \Big(\dfrac{-3}{2}\Big)^{-4 \times \dfrac{1}{2}}\\[1em] \Rightarrow \Big(\dfrac{-3}{2}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{2}{-3}\Big)^{2}\\[1em] \Rightarrow \dfrac{2 \times 2}{(-3) \times (-3)}\\[1em] \Rightarrow \dfrac{4}{9}$

Hence, the multiplicative inverse of $\mathbf{\Big\lbrace\Big(\dfrac{-3}{2}\Big)^{-4}\Big\rbrace^{\dfrac{1}{2}}}$ is $\mathbf{\dfrac{9}{4}}$.

Find the multiplicative inverse of:

$\Big(\dfrac{5}{7}\Big)^{-2} \times \Big(\dfrac{5}{7}\Big)^{4} \div \Big(\dfrac{5}{7}\Big)^{3}$

Answer

According to the laws of exponents,

$a^m \times a^n = a^{m+n}$ and $a^m \div a^n = a^{m-n}$

$\Rightarrow \Big(\dfrac{5}{7}\Big)^{-2} \times \Big(\dfrac{5}{7}\Big)^{4} \div \Big(\dfrac{5}{7}\Big)^{3}\\[1em] \Rightarrow \Big(\dfrac{5}{7}\Big)^{-2 + 4 - 3}\\[1em] \Rightarrow \Big(\dfrac{5}{7}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{7}{5}\Big)^{1}\\[1em] \Rightarrow \dfrac{7}{5}$

Hence, the multiplicative inverse of $\mathbf{\Big(\dfrac{5}{7}\Big)^{-2} \times \Big(\dfrac{5}{7}\Big)^{4} \div \Big(\dfrac{5}{7}\Big)^{3}}$ is $\mathbf{\dfrac{5}{7}}$.

Express 16^-2 as a power with base 2.

Answer

We know that,

16 = 2 x 2 x 2 x 2 = 2⁴

Solving,

⇒ 16^-2

⇒ (2⁴)^-2

⇒ 2^{4 × (-2)}

⇒ 2^-8

Hence, 16^-2 = 2^-8.

Express 125^-4 as a power with base 5.

Answer

We know that 125 = 5 x 5 x 5 = 5³

⇒ 125^-4

⇒ (5³)^-4

⇒ 5^{3 × (-4)}

⇒ 5^-12

Hence, 125^-4 = 5^-12.

Simplify and write in exponential form with positive exponent:

$\Big[\Big\lbrace\Big(\dfrac{5}{7}\Big)^2\Big\rbrace^{-1}\Big]^{-3}$

Answer

Solving,

$\Rightarrow \Big[\Big\lbrace\Big(\dfrac{5}{7}\Big)^2\Big\rbrace^{-1}\Big]^{-3}\\[1em] \Rightarrow \Big(\dfrac{5}{7}\Big)^{2 \times (-1) \times (-3)}\\[1em] \Rightarrow \Big(\dfrac{5}{7}\Big)^{6}$

Hence, $\mathbf{\Big[\Big\lbrace\Big(\dfrac{5}{7}\Big)^2\Big\rbrace^{-1}\Big]^{-3} = \Big(\dfrac{5}{7}\Big)^{6}}$.

Simplify and write in exponential form with positive exponent:

$\Big(\dfrac{2}{7}\Big)^2 \times \Big(\dfrac{7}{2}\Big)^{-3} \div \Big\lbrace\Big(\dfrac{7}{5}\Big)^{-2}\Big\rbrace^{-4}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{2}{7}\Big)^2 \times \Big(\dfrac{7}{2}\Big)^{-3} \div \Big\lbrace\Big(\dfrac{7}{5}\Big)^{-2}\Big\rbrace^{-4}\\[1em] \Rightarrow \Big(\dfrac{2}{7}\Big)^2 \times \Big(\dfrac{2}{7}\Big)^{3} \div \Big(\dfrac{7}{5}\Big)^{8}\\[1em] \Rightarrow \Big(\dfrac{2}{7}\Big)^{2 + 3} \div \Big(\dfrac{7}{5}\Big)^{8}\\[1em] \Rightarrow \Big(\dfrac{2}{7}\Big)^{5} \times \Big(\dfrac{5}{7}\Big)^{8}\\[1em] \Rightarrow \dfrac{2^5}{7^5} \times \dfrac{5^8}{7^8}\\[1em] \Rightarrow \dfrac{2^5 \times 5^8}{7^{5+8}}\\[1em] \Rightarrow \dfrac{2^5 \times 5^8}{7^{13}}$

Hence, $\mathbf{\Big(\dfrac{2}{7}\Big)^2 \times \Big(\dfrac{7}{2}\Big)^{-3} \div \Big\lbrace\Big(\dfrac{7}{5}\Big)^{-2}\Big\rbrace^{-4} = \dfrac{2^5 \times 5^8}{7^{13}}}$.

Simplify and write in exponential form with positive exponent:

$\Big(\dfrac{4}{5}\Big)^2 \times 5^4 \times \Big(\dfrac{2}{5}\Big)^{-2} \div \Big(\dfrac{5}{2}\Big)^{-3}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{4}{5}\Big)^2 \times 5^4 \times \Big(\dfrac{2}{5}\Big)^{-2} \div \Big(\dfrac{5}{2}\Big)^{-3}\\[1em] \Rightarrow \dfrac{2^4}{5^2} \times 5^4 \times \Big(\dfrac{5}{2}\Big)^{2} \div \Big(\dfrac{2}{5}\Big)^{3}\\[1em] \Rightarrow \dfrac{2^4}{5^2} \times 5^4 \times \dfrac{5^2}{2^2} \times \dfrac{5^3}{2^3}\\[1em] \Rightarrow \dfrac{2^4 \times 5^4 \times 5^2 \times 5^3}{5^2 \times 2^2 \times 2^3}\\[1em] \Rightarrow 2^{4 - 2 - 3} \times 5^{-2 + 4 + 2 + 3}\\[1em] \Rightarrow 2^{-1} \times 5^{7}\\[1em] \Rightarrow \dfrac{5^7}{2^1}$

Hence, $\mathbf{\Big(\dfrac{4}{5}\Big)^2 \times 5^4 \times \Big(\dfrac{2}{5}\Big)^{-2} \div \Big(\dfrac{5}{2}\Big)^{-3} = \dfrac{5^7}{2^1}}$.

Simplify and write in exponential form with positive exponent:

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

Answer

Solving,

We know that 8 = 2³

$\Rightarrow \dfrac{8^{-1} \times 5^3}{2^{-4}}\\[1em] \Rightarrow \dfrac{(2^3)^{-1} \times 5^3}{2^{-4}}\\[1em] \Rightarrow \dfrac{2^{-3} \times 5^3}{2^{-4}}\\[1em] \Rightarrow 2^{-3 - (-4)} \times 5^3\\[1em] \Rightarrow 2^{-3 + 4} \times 5^3\\[1em] \Rightarrow 2^{1} \times 5^{3}\\[1em] \Rightarrow 5^3 \times 2^1$

Hence, $\mathbf{\dfrac{8^{-1} \times 5^3}{2^{-4}} = 5^3 \times 2^1}$.

Simplify and write the following in exponential form:

$((-2)^3)^2 + 5^{-3} \div 5^{-5} - \Big(-\dfrac{1}{2}\Big)^0$

Answer

Simplifying,

$\Rightarrow ((-2)^3)^2 + 5^{-3} \div 5^{-5} - \Big(-\dfrac{1}{2}\Big)^0\\[1em] \Rightarrow (-2)^{3 \times 2} + 5^{-3 - (-5)} - 1\\[1em] \Rightarrow (-2)^{6} + 5^{-3 + 5} - 1\\[1em] \Rightarrow 64 + 5^{2} - 1\\[1em] \Rightarrow 64 + 25 - 1\\[1em] \Rightarrow 88\\[1em] \Rightarrow 2 \times 2 \times 2 \times 11\\[1em] \Rightarrow 2^{3} \times 11$

Hence, $\mathbf{((-2)^3)^2 + 5^{-3} \div 5^{-5} - \Big(-\dfrac{1}{2}\Big)^0 = 2^3 \times 11}$.

Simplify and write the following in exponential form:

$3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \Big(\dfrac{2}{3}\Big)^{-1} + 2^{-1} + \Big(\dfrac{1}{19}\Big)^{-1}$

Answer

Simplifying,

$\Rightarrow 3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \Big(\dfrac{2}{3}\Big)^{-1} + 2^{-1} + \Big(\dfrac{1}{19}\Big)^{-1}\\[1em] \Rightarrow 3^{-5 + 2 - (-6)} + (2^2)^2 \times 3^2 + \dfrac{3}{2} + \dfrac{1}{2} + 19\\[1em] \Rightarrow 3^{-5 + 2 + 6} + 2^4 \times 3^2 + \dfrac{3}{2} + \dfrac{1}{2} + 19\\[1em] \Rightarrow 3^{3} + (16 \times 9) + \dfrac{3 + 1}{2} + 19\\[1em] \Rightarrow 27 + 144 + \dfrac{4}{2} + 19\\[1em] \Rightarrow 27 + 144 + 2 + 19\\[1em] \Rightarrow 192\\[1em] \Rightarrow 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow 2^6 \times 3^1$

Hence, $\mathbf{3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \Big(\dfrac{2}{3}\Big)^{-1} + 2^{-1} + \Big(\dfrac{1}{19}\Big)^{-1} = 2^6 \times 3^1}$.

Simplify and write in exponential form with negative exponent:

$5^3 \times \Big(\dfrac{4}{5}\Big)^3$

Answer

Simplifying,

$\Rightarrow 5^3 \times \Big(\dfrac{4}{5}\Big)^3\\[1em] \Rightarrow 5^3 \times \dfrac{4^3}{5^3}\\[1em] \Rightarrow \dfrac{5^3 \times 4^3}{5^3}\\[1em] \Rightarrow 4^3\\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big)^{-3}$

Hence, $\mathbf{5^3 \times \Big(\dfrac{4}{5}\Big)^3 = \Big(\dfrac{1}{4}\Big)^{-3}}$.

Simplify and write in exponential form with negative exponent:

$\Big[\Big(\dfrac{3}{7}\Big)^{-2}\Big]^{-3}$

Answer

Simplifying,

$\Rightarrow \Big[\Big(\dfrac{3}{7}\Big)^{-2}\Big]^{-3}\\[1em] \Rightarrow \Big(\dfrac{3}{7}\Big)^{(-2) \times (-3)}\\[1em] \Rightarrow \Big(\dfrac{3}{7}\Big)^{6}\\[1em] \Rightarrow \Big(\dfrac{7}{3}\Big)^{-6}$

Hence, $\mathbf{\Big[\Big(\dfrac{3}{7}\Big)^{-2}\Big]^{-3} = \Big(\dfrac{7}{3}\Big)^{-6}}$.

Simplify and write in exponential form with negative exponent:

$\Big(\dfrac{5}{9}\Big)^{-2} \times \Big(\dfrac{5}{3}\Big)^2 \div \Big(\dfrac{1}{5}\Big)^{-2}$

Answer

Simplifying,

$\Rightarrow \Big(\dfrac{5}{9}\Big)^{-2} \times \Big(\dfrac{5}{3}\Big)^2 \div \Big(\dfrac{1}{5}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{9}{5}\Big)^{2} \times \Big(\dfrac{5}{3}\Big)^2 \div (5)^{2}\\[1em] \Rightarrow \dfrac{3^4}{5^2} \times \dfrac{5^2}{3^2} \times \dfrac{1}{5^2}\\[1em] \Rightarrow 3^{4 - 2} \times 5^{-2 + 2 - 2}\\[1em] \Rightarrow 3^{2} \times 5^{-2}\\[1em] \Rightarrow \dfrac{3^2}{5^2}\\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{2}\\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{-2}$

Hence, $\mathbf{\Big(\dfrac{5}{9}\Big)^{-2} \times \Big(\dfrac{5}{3}\Big)^2 \div \Big(\dfrac{1}{5}\Big)^{-2} = \Big(\dfrac{5}{3}\Big)^{-2}}$.

Simplify and write in exponential form with negative exponent:

$2^{-1}\Big[\Big(\dfrac{5}{3}\Big)^4 + \Big(\dfrac{3}{5}\Big)^{-2}\Big] \div \dfrac{17}{9}$

Answer

Simplifying,

$\Rightarrow 2^{-1}\Big[\Big(\dfrac{5}{3}\Big)^4 + \Big(\dfrac{3}{5}\Big)^{-2}\Big] \div \dfrac{17}{9}\\[1em] \Rightarrow \dfrac{1}{2}\Big[\Big(\dfrac{5}{3}\Big)^4 + \Big(\dfrac{5}{3}\Big)^{2}\Big] \div \dfrac{17}{9}\\[1em] \Rightarrow \dfrac{1}{2}\Big[\dfrac{625}{81} + \dfrac{25}{9}\Big] \div \dfrac{17}{9}$

LCM of 81 and 9 is 81

$\Rightarrow \dfrac{1}{2}\Big[\dfrac{625}{81} + \dfrac{25 \times 9}{9 \times 9}\Big] \div \dfrac{17}{9}\\[1em] \Rightarrow \dfrac{1}{2}\Big[\dfrac{625}{81} + \dfrac{225}{81}\Big] \div \dfrac{17}{9}\\[1em] \Rightarrow \dfrac{1}{2}\Big[\dfrac{625 + 225}{81}\Big] \div \dfrac{17}{9}\\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{850}{81} \times \dfrac{9}{17}\\[1em] \Rightarrow \dfrac{1 \times 850 \times 9}{2 \times 81 \times 17}\\[1em] \Rightarrow \dfrac{7650}{2754}\\[1em] \Rightarrow \dfrac{25}{9}\\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{2}\\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{-2}$

Hence, $\mathbf{2^{-1}\Big[\Big(\dfrac{5}{3}\Big)^4 + \Big(\dfrac{3}{5}\Big)^{-2}\Big] \div \dfrac{17}{9} = \Big(\dfrac{3}{5}\Big)^{-2}}$.

Simplify and write in exponential form with negative exponent:

$(-7)^3 \times \Big(\dfrac{1}{-7}\Big)^{-9} \div (-7)^{10}$

Answer

Simplifying,

$\Rightarrow (-7)^3 \times \Big(\dfrac{1}{-7}\Big)^{-9} \div (-7)^{10}\\[1em] \Rightarrow (-7)^{3} \times (-7)^{9} \div (-7)^{10}\\[1em] \Rightarrow (-7)^{3 + 9 - 10}\\[1em] \Rightarrow (-7)^{2}\\[1em] \Rightarrow 7^{2}\\[1em] \Rightarrow \Big(\dfrac{1}{7}\Big)^{-2}$

Hence, $\mathbf{(-7)^3 \times \Big(\dfrac{1}{-7}\Big)^{-9} \div (-7)^{10} = \Big(\dfrac{1}{7}\Big)^{-2}}$.

Simplify:

$\dfrac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}$ $(z \neq 0)$

Answer

We know that,

49 = 7²

Simplifying,

$\Rightarrow \dfrac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}\\[1em] \Rightarrow \dfrac{7^2 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}\\[1em] \Rightarrow \dfrac{7^{2 - (-3)} \times z^{-3 - (-5)}}{10}\\[1em] \Rightarrow \dfrac{7^{2 + 3} \times z^{-3 + 5}}{10}\\[1em] \Rightarrow \dfrac{7^{5} \times z^{2}}{10}$

Hence, $\mathbf{\dfrac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} = \dfrac{7^{5} \times z^{2}}{10}}$.

Simplify:

$\dfrac{9^3 \times 27 \times t^4}{(3)^2 \times (3)^4 \times t^2}$

Answer

We know that,

9 = 3² and 27 = 3³

Simplifying,

$\Rightarrow \dfrac{9^3 \times 27 \times t^4}{(3)^2 \times (3)^4 \times t^2}\\[1em] \Rightarrow \dfrac{(3^2)^3 \times 3^3 \times t^4}{3^2 \times 3^4 \times t^2}\\[1em] \Rightarrow \dfrac{3^6 \times 3^3 \times t^4}{3^2 \times 3^4 \times t^2}\\[1em] \Rightarrow \dfrac{3^{6 + 3} \times t^4}{3^{2 + 4} \times t^2}\\[1em] \Rightarrow \dfrac{3^{9} \times t^4}{3^{6} \times t^2}\\[1em] \Rightarrow 3^{9 - 6} \times t^{4 - 2}\\[1em] \Rightarrow 3^{3} \times t^{2}\\[1em] \Rightarrow 27,t^2$

Hence, $\mathbf{\dfrac{9^3 \times 27 \times t^4}{(3)^2 \times (3)^4 \times t^2} = 27,t^2}$.

Simplify:

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

Answer

Simplifying,

$\Rightarrow \dfrac{(3^{-2})^2 \times (5^2)^{-3} \times (t^{-3})^2}{(3^{-2})^5 \times (5^3)^{-2} \times (t^{-4})^3}\\[1em] \Rightarrow \dfrac{3^{-4} \times 5^{-6} \times t^{-6}}{3^{-10} \times 5^{-6} \times t^{-12}}\\[1em] \Rightarrow 3^{-4 - (-10)} \times 5^{-6 - (-6)} \times t^{-6 - (-12)}\\[1em] \Rightarrow 3^{-4 + 10} \times 5^{-6 + 6} \times t^{-6 + 12}\\[1em] \Rightarrow 3^{6} \times 5^{0} \times t^{6}\\[1em] \Rightarrow 3^{6} \times 1 \times t^{6}\\[1em] \Rightarrow 3^{6} \times t^{6}$

Hence, $\mathbf{\dfrac{(3^{-2})^2 \times (5^2)^{-3} \times (t^{-3})^2}{(3^{-2})^5 \times (5^3)^{-2} \times (t^{-4})^3} = 3^{6} \times t^{6}}$.

Simplify:

$\dfrac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}}$

Answer

We know that,

15 = 3 x 5, 500 = 2² x 5³ and 6 = 2 x 3

Simplifying,

$\Rightarrow \dfrac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}}\\[1em] \Rightarrow \dfrac{2^{-5} \times (3 \times 5)^{-5} \times (2^2 \times 5^3)}{5^{-6} \times (2 \times 3)^{-5}}\\[1em] \Rightarrow \dfrac{2^{-5} \times 3^{-5} \times 5^{-5} \times 2^2 \times 5^3}{5^{-6} \times 2^{-5} \times 3^{-5}}\\[1em] \Rightarrow \dfrac{2^{-5 + 2} \times 3^{-5} \times 5^{-5 + 3}}{2^{-5} \times 3^{-5} \times 5^{-6}}\\[1em] \Rightarrow \dfrac{2^{-3} \times 3^{-5} \times 5^{-2}}{2^{-5} \times 3^{-5} \times 5^{-6}}\\[1em] \Rightarrow 2^{-3 - (-5)} \times 3^{-5 - (-5)} \times 5^{-2 - (-6)}\\[1em] \Rightarrow 2^{-3 + 5} \times 3^{0} \times 5^{-2 + 6}\\[1em] \Rightarrow 2^{2} \times 1 \times 5^{4}\\[1em] \Rightarrow 4 \times 625\\[1em] \Rightarrow 2500$

Hence, $\mathbf{\dfrac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}} = 2500}$.

By what number should $\Big(\dfrac{3}{-2}\Big)^{-3}$ be divided to get $\Big(\dfrac{2}{3}\Big)^2$?

Answer

Let the required number be $x$. So,

$\Rightarrow \Big(\dfrac{3}{-2}\Big)^{-3} \div x = \Big(\dfrac{2}{3}\Big)^2\\[1em] \Rightarrow \Big(\dfrac{-2}{3}\Big)^{3} \div x = \Big(\dfrac{2}{3}\Big)^2\\[1em] \Rightarrow \dfrac{-8}{27} \div x = \dfrac{4}{9}\\[1em] \Rightarrow \dfrac{-8}{27} \times \dfrac{1}{x} = \dfrac{4}{9}\\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{4}{9} \times \dfrac{27}{-8}\\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{4 \times 27}{9 \times (-8)}\\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{108}{-72}\\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-3}{2}\\[1em] \Rightarrow x = \dfrac{-2}{3}$

Hence, $\mathbf{\Big(\dfrac{3}{-2}\Big)^{-3}}$ should be divided by $\mathbf{\dfrac{-2}{3}}$ to get $\mathbf{\Big(\dfrac{2}{3}\Big)^2}$.

Find the value of m for which 9^m ÷ 3^-2 = 9⁴

Answer

We know that 9 = 3²

$\Rightarrow 9^m \div 3^{-2} = 9^4\\[1em] \Rightarrow (3^2)^m \div 3^{-2} = (3^2)^4\\[1em] \Rightarrow 3^{2m} \div 3^{-2} = 3^{8}\\[1em] \Rightarrow 3^{2m - (-2)} = 3^{8}\\[1em] \Rightarrow 3^{2m + 2} = 3^{8}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2m + 2 = 8\\[1em] \Rightarrow 2m = 8 - 2\\[1em] \Rightarrow 2m = 6\\[1em] \Rightarrow m = \dfrac{6}{2}\\[1em] \Rightarrow m = 3$

Hence, the value of m is 3.

If $\Big(\dfrac{-5}{7}\Big)^{-4} \times \Big(\dfrac{-5}{7}\Big)^{12} = \Big\lbrace\Big(\dfrac{-5}{7}\Big)^3\Big\rbrace^x \times \Big(\dfrac{-5}{7}\Big)^{-1}$, find the value of x.

Answer

Solving,

$\Rightarrow \Big(\dfrac{-5}{7}\Big)^{-4} \times \Big(\dfrac{-5}{7}\Big)^{12} = \Big\lbrace\Big(\dfrac{-5}{7}\Big)^3\Big\rbrace^x \times \Big(\dfrac{-5}{7}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{-5}{7}\Big)^{-4 + 12} = \Big(\dfrac{-5}{7}\Big)^{3x} \times \Big(\dfrac{-5}{7}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{-5}{7}\Big)^{8} = \Big(\dfrac{-5}{7}\Big)^{3x - 1}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 8 = 3x - 1\\[1em] \Rightarrow 8 + 1 = 3x\\[1em] \Rightarrow 9 = 3x\\[1em] \Rightarrow x = \dfrac{9}{3}\\[1em] \Rightarrow x = 3$

Hence, the value of x is 3.

Find x if $\Big(\dfrac{-2}{3}\Big)^{-13} \times \Big(\dfrac{3}{-2}\Big)^{8} = \Big(\dfrac{-2}{3}\Big)^{-2x+1}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{-2}{3}\Big)^{-13} \times \Big(\dfrac{3}{-2}\Big)^{8} = \Big(\dfrac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\dfrac{-2}{3}\Big)^{-13} \times \Big(\dfrac{-2}{3}\Big)^{-8} = \Big(\dfrac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\dfrac{-2}{3}\Big)^{-13 + (-8)} = \Big(\dfrac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\dfrac{-2}{3}\Big)^{-21} = \Big(\dfrac{-2}{3}\Big)^{-2x+1}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow -21 = -2x + 1\\[1em] \Rightarrow -21 - 1 = -2x\\[1em] \Rightarrow -22 = -2x\\[1em] \Rightarrow x = \dfrac{-22}{-2}\\[1em] \Rightarrow x = 11$

Hence, the value of x is 11.

If $5^{2x-1} = \dfrac{1}{(125)^{x-3}}$, find x

Answer

We know that 125 = 5³

$\Rightarrow 5^{2x-1} = \dfrac{1}{(125)^{x-3}}\\[1em] \Rightarrow 5^{2x-1} = \dfrac{1}{(5^3)^{x-3}}\\[1em] \Rightarrow 5^{2x-1} = \dfrac{1}{5^{3(x-3)}}\\[1em] \Rightarrow 5^{2x-1} = 5^{-3(x-3)}\\[1em] \Rightarrow 5^{2x-1} = 5^{-3x+9}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2x - 1 = -3x + 9\\[1em] \Rightarrow 2x + 3x = 9 + 1\\[1em] \Rightarrow 5x = 10\\[1em] \Rightarrow x = \dfrac{10}{5}\\[1em] \Rightarrow x = 2$

Hence, the value of x is 2.

If $\dfrac{9^n \times 3^5 \times (27)^3}{3 \times (81)^4} = 27$, find n

Answer

We know that 9 = 3², 27 = 3³ and 81 = 3⁴

$\Rightarrow \dfrac{9^n \times 3^5 \times (27)^3}{3 \times (81)^4} = 27\\[1em] \Rightarrow \dfrac{(3^2)^n \times 3^5 \times (3^3)^3}{3 \times (3^4)^4} = 3^3\\[1em] \Rightarrow \dfrac{3^{2n} \times 3^5 \times 3^{9}}{3^1 \times 3^{16}} = 3^3\\[1em] \Rightarrow \dfrac{3^{2n + 5 + 9}}{3^{1 + 16}} = 3^3\\[1em] \Rightarrow \dfrac{3^{2n + 14}}{3^{17}} = 3^3\\[1em] \Rightarrow 3^{2n + 14 - 17} = 3^3\\[1em] \Rightarrow 3^{2n - 3} = 3^3$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2n - 3 = 3\\[1em] \Rightarrow 2n = 3 + 3\\[1em] \Rightarrow 2n = 6\\[1em] \Rightarrow n = \dfrac{6}{2}\\[1em] \Rightarrow n = 3$

Hence, the value of n is 3.