Exponents and Powers — Check Your Progress

Evaluate:

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

Answer

Solving,

$\Rightarrow \Big(\dfrac{5}{4}\Big)^{-3}\\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^{3}\\[1em] \Rightarrow \dfrac{4 \times 4 \times 4}{5 \times 5 \times 5}\\[1em] \Rightarrow \dfrac{64}{125}$

Hence, $\mathbf{\Big(\dfrac{5}{4}\Big)^{-3} = \dfrac{64}{125}}$.

Evaluate:

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

Answer

Solving,

$\Rightarrow \Big(\dfrac{-3}{7}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{7}{-3}\Big)^{2}\\[1em] \Rightarrow \dfrac{7 \times 7}{(-3) \times (-3)}\\[1em] \Rightarrow \dfrac{49}{9}\\[1em] \Rightarrow 5\dfrac{4}{9}$

Hence, $\mathbf{\Big(\dfrac{-3}{7}\Big)^{-2} = 5\dfrac{4}{9}}$.

Simplify:

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

Answer

Solving,

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

⇒ [2³ + 3³ + 4³] ÷ 3²

⇒ [8 + 27 + 64] ÷ 9

⇒ 99 ÷ 9

$\Rightarrow \dfrac{99}{9}$

⇒ 11

Hence, the value of the given expression is 11.

Simplify:

$\Big[3^3 - \Big(\dfrac{1}{2}\Big)^{-3}\Big] \times \dfrac{1}{19}$

Answer

Solving,

$\Rightarrow \Big[3^3 - \Big(\dfrac{1}{2}\Big)^{-3}\Big] \times \dfrac{1}{19}\\[1em] \Rightarrow \Big[3^3 - 2^3\Big] \times \dfrac{1}{19}\\[1em] \Rightarrow [27 - 8] \times \dfrac{1}{19}\\[1em] \Rightarrow 19 \times \dfrac{1}{19}\\[1em] \Rightarrow \dfrac{19}{19}\\[1em] \Rightarrow 1$

Hence, $\mathbf{\Big[3^3 - \Big(\dfrac{1}{2}\Big)^{-3}\Big] \times \dfrac{1}{19} = 1}$.

Simplify:

$[2^{-1} + 3^0 + 5^1 + 7^2 + 9^3] \div \Big(\dfrac{2}{3}\Big)^{-1}$

Answer

Solving,

$\Rightarrow [2^{-1} + 3^0 + 5^1 + 7^2 + 9^3] \div \Big(\dfrac{2}{3}\Big)^{-1}\\[1em] \Rightarrow \Big[\dfrac{1}{2} + 1 + 5 + 49 + 729\Big] \div \Big(\dfrac{3}{2}\Big)\\[1em] \Rightarrow \Big[\dfrac{1}{2} + 784\Big] \div \Big(\dfrac{3}{2}\Big)\\[1em] \Rightarrow \Big[\dfrac{1 + 1568}{2}\Big] \div \Big(\dfrac{3}{2}\Big)\\[1em] \Rightarrow \dfrac{1569}{2} \times \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{1569 \times 2}{2 \times 3}\\[1em] \Rightarrow \dfrac{1569}{3}\\[1em] \Rightarrow 523$

Hence, $\mathbf{[2^{-1} + 3^0 + 5^1 + 7^2 + 9^3] \div \Big(\dfrac{2}{3}\Big)^{-1} = 523}$.

Simplify:

(7^-1 - 8^-1)^-1 - (3^-1 - 4^-1)^-1

Answer

Solving,

$\Rightarrow (7^{-1} - 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{7} - \dfrac{1}{8}\Big)^{-1} - \Big(\dfrac{1}{3} - \dfrac{1}{4}\Big)^{-1}$

LCM of 7 and 8 is 56, and LCM of 3 and 4 is 12

$\Rightarrow \Big(\dfrac{8}{56} - \dfrac{7}{56}\Big)^{-1} - \Big(\dfrac{4}{12} - \dfrac{3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{8 - 7}{56}\Big)^{-1} - \Big(\dfrac{4 - 3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{56}\Big)^{-1} - \Big(\dfrac{1}{12}\Big)^{-1}\\[1em] \Rightarrow 56 - 12\\[1em] \Rightarrow 44$

Hence, $(7^{-1} - 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1} = 44$.

Find the multiplicative inverse of $\Big(\dfrac{5}{3}\Big)^{-2}$

Answer

Solving,

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

Hence, the multiplicative inverse of $\mathbf{\Big(\dfrac{5}{3}\Big)^{-2}}$ is $\mathbf{\dfrac{25}{9}}$.

Simplify and write in exponential form:

$\dfrac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}$

Answer

We know that 15 = 3 x 5, 625 = 5⁴ and 10 = 2 x 5

$\Rightarrow \dfrac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}\\[1em] \Rightarrow \dfrac{2^{-4} \times (3 \times 5)^{-3} \times 5^4}{5^2 \times (2 \times 5)^{-4}}\\[1em] \Rightarrow \dfrac{2^{-4} \times 3^{-3} \times 5^{-3} \times 5^4}{5^2 \times 2^{-4} \times 5^{-4}}\\[1em] \Rightarrow \dfrac{2^{-4} \times 3^{-3} \times 5^{-3 + 4}}{2^{-4} \times 5^{2 - 4}}\\[1em] \Rightarrow \dfrac{2^{-4} \times 3^{-3} \times 5^{1}}{2^{-4} \times 5^{-2}}\\[1em] \Rightarrow 2^{-4 - (-4)} \times 3^{-3} \times 5^{1 - (-2)}\\[1em] \Rightarrow 2^{0} \times 3^{-3} \times 5^{3}\\[1em] \Rightarrow \dfrac{5^3}{3^3}\\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{3}$

Hence, $\mathbf{\dfrac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}} = \Big(\dfrac{5}{3}\Big)^{3}}$.

Simplify and write in exponential form:

$\Big(\dfrac{9}{4}\Big)^{-1} \times \Big(\dfrac{2}{3}\Big)^2 \div \Big(\dfrac{3}{2}\Big)^{-2}$

Answer

Solving,

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

Hence, $\mathbf{\Big(\dfrac{9}{4}\Big)^{-1} \times \Big(\dfrac{2}{3}\Big)^2 \div \Big(\dfrac{3}{2}\Big)^{-2} = \dfrac{4}{9}}$.

Simplify: $\dfrac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}}$

Answer

Solving,

$\Rightarrow \dfrac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}}\\[1em] \Rightarrow \dfrac{5^{n} \times 5^2 - 6 \times 5^{n} \times 5}{13 \times 5^n - 2 \times 5^{n} \times 5}\\[1em] \Rightarrow \dfrac{5^{n}(5^2 - 6 \times 5)}{5^n(13 - 2 \times 5)}\\[1em] \Rightarrow \dfrac{5^{n}(25 - 30)}{5^n(13 - 10)}\\[1em] \Rightarrow \dfrac{25 - 30}{13 - 10}\\[1em] \Rightarrow \dfrac{-5}{3}$

Hence, $\mathbf{\dfrac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}} = \dfrac{-5}{3}}$.

Prove that: (a + b)^-1 (a^-1 + b^-1) = (ab)^-1

Answer

To prove: $(a + b)^{-1}(a^{-1} + b^{-1}) = (ab)^{-1}$

Taking L.H.S.,

$\Rightarrow (a + b)^{-1}(a^{-1} + b^{-1})\\[1em] \Rightarrow \dfrac{1}{a + b}\Big(\dfrac{1}{a} + \dfrac{1}{b}\Big)\\[1em] \Rightarrow \dfrac{1}{a + b}\Big(\dfrac{b + a}{ab}\Big)\\[1em] \Rightarrow \dfrac{1}{a + b} \times \dfrac{a + b}{ab}\\[1em] \Rightarrow \dfrac{a + b}{(a + b) \times ab}\\[1em] \Rightarrow \dfrac{1}{ab}\\[1em] \Rightarrow (ab)^{-1}\\[1em] \Rightarrow \text{R.H.S.}$

Since, L.H.S. = R.H.S.

Hence, $\bm{(a + b)^{-1}(a^{-1} + b^{-1}) = (ab)^{-1}}$ is proved.

Find the value of x for which $\Big\lbrace\Big(\dfrac{-2}{7}\Big)^2\Big\rbrace^x \times \Big(\dfrac{-7}{2}\Big)^{-1} = \dfrac{-8}{343}$

Answer

We know that $\dfrac{-8}{343} = \dfrac{(-2)^3}{7^3} = \Big(\dfrac{-2}{7}\Big)^3$

$\Rightarrow \Big\lbrace\Big(\dfrac{-2}{7}\Big)^2\Big\rbrace^x \times \Big(\dfrac{-7}{2}\Big)^{-1} = \dfrac{-8}{343}\\[1em] \Rightarrow \Big(\dfrac{-2}{7}\Big)^{2x} \times \Big(\dfrac{-2}{7}\Big)^{1} = \Big(\dfrac{-2}{7}\Big)^{3}\\[1em] \Rightarrow \Big(\dfrac{-2}{7}\Big)^{2x + 1} = \Big(\dfrac{-2}{7}\Big)^{3}$

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

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

Hence, the value of x is 1.

If $\dfrac{2^{-n} \times 8^{2n+1} \times 16^{2n}}{4^{3n}} = \dfrac{1}{16}$, find the value of n.

Answer

We know that 8 = 2³, 16 = 2⁴ and 4 = 2²

$\Rightarrow \dfrac{2^{-n} \times 8^{2n+1} \times 16^{2n}}{4^{3n}} = \dfrac{1}{16}\\[1em] \Rightarrow \dfrac{2^{-n} \times (2^3)^{2n+1} \times (2^4)^{2n}}{(2^2)^{3n}} = \dfrac{1}{2^4}\\[1em] \Rightarrow \dfrac{2^{-n} \times 2^{6n+3} \times 2^{8n}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow \dfrac{2^{-n + 6n + 3 + 8n}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow \dfrac{2^{13n + 3}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow 2^{13n + 3 - 6n} = 2^{-4}\\[1em] \Rightarrow 2^{7n + 3} = 2^{-4}$

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

$\Rightarrow 7n + 3 = -4\\[1em] \Rightarrow 7n = -4 - 3\\[1em] \Rightarrow 7n = -7\\[1em] \Rightarrow n = \dfrac{-7}{7}\\[1em] \Rightarrow n = -1$

Hence, the value of n is -1.

By what number should $\Big[\Big(\dfrac{-5}{2}\Big)^3\Big]^{-3}$ be multiplied to get $\Big(\dfrac{-2}{5}\Big)^5$?

Answer

Let the required number be $x$. So,

$\Rightarrow \Big[\Big(\dfrac{-5}{2}\Big)^3\Big]^{-3} \times x = \Big(\dfrac{-2}{5}\Big)^5\\[1em] \Rightarrow \Big(\dfrac{-5}{2}\Big)^{-9} \times x = \Big(\dfrac{-2}{5}\Big)^5\\[1em] \Rightarrow \Big(\dfrac{-2}{5}\Big)^{9} \times x = \Big(\dfrac{-2}{5}\Big)^5\\[1em] \Rightarrow x = \Big(\dfrac{-2}{5}\Big)^5 \div \Big(\dfrac{-2}{5}\Big)^9\\[1em] \Rightarrow x = \Big(\dfrac{-2}{5}\Big)^{5 - 9}\\[1em] \Rightarrow x = \Big(\dfrac{-2}{5}\Big)^{-4}\\[1em] \Rightarrow x = \Big(\dfrac{-5}{2}\Big)^{4}\\[1em] \Rightarrow x = \dfrac{(-5)^4}{2^4}\\[1em] \Rightarrow x = \dfrac{625}{16}$

Hence, $\mathbf{\Big[\Big(\dfrac{-5}{2}\Big)^3\Big]^{-3}}$ should be multiplied by $\mathbf{\dfrac{625}{16}}$ to get $\mathbf{\Big(\dfrac{-2}{5}\Big)^5}$.

By what number should $\Big(\dfrac{-3}{2}\Big)^{-4}$ be divided to get $\Big(\dfrac{-2}{3}\Big)^3$?

Answer

Let the required number be $x$. So,

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

Hence, $\mathbf{\Big(\dfrac{-3}{2}\Big)^{-4}}$ should be divided by $\mathbf{\dfrac{-2}{3}}$ to get $\mathbf{\Big(\dfrac{-2}{3}\Big)^3}$.

If $\Big(\dfrac{9}{4}\Big)^{-4} \times \Big(\dfrac{2}{3}\Big)^3 = \Big(\dfrac{p}{q}\Big)^{11}$, then find the value of $\Big(\dfrac{p}{q}\Big)^{-2}$

Answer

We know that 9 = 3² and 4 = 2²

$\Rightarrow \Big(\dfrac{9}{4}\Big)^{-4} \times \Big(\dfrac{2}{3}\Big)^3 = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \Big(\dfrac{4}{9}\Big)^{4} \times \Big(\dfrac{2}{3}\Big)^3 = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \Big(\dfrac{2^2}{3^2}\Big)^{4} \times \dfrac{2^3}{3^3} = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \dfrac{2^8}{3^8} \times \dfrac{2^3}{3^3} = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \dfrac{2^{8 + 3}}{3^{8 + 3}} = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \dfrac{2^{11}}{3^{11}} = \Big(\dfrac{p}{q}\Big)^{11}\\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{11} = \Big(\dfrac{p}{q}\Big)^{11}$

So, $\dfrac{p}{q} = \dfrac{2}{3}$

Now,

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

Hence, $\mathbf{\Big(\dfrac{p}{q}\Big)^{-2} = \dfrac{9}{4}}$.

Find the multiplicative inverse of (5⁰ + 3⁰)(5⁰ - 3⁰)

Answer

We know that $a^0 = 1$

⇒ (5⁰ + 3⁰)(5⁰ - 3⁰)

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0

Since the multiplicative inverse of a number is obtained by taking its reciprocal, and the reciprocal of 0 is not defined, the multiplicative inverse of 0 does not exist.

Hence, the multiplicative inverse of (5⁰ + 3⁰)(5⁰ - 3⁰) does not exist.

Evaluate $\Big(\dfrac{x^p}{x^q}\Big)^{p+q} \times \Big(\dfrac{x^q}{x^r}\Big)^{q+r} \times \Big(\dfrac{x^r}{x^p}\Big)^{r+p}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{x^p}{x^q}\Big)^{p+q} \times \Big(\dfrac{x^q}{x^r}\Big)^{q+r} \times \Big(\dfrac{x^r}{x^p}\Big)^{r+p}\\[1em] \Rightarrow (x^{p-q})^{p+q} \times (x^{q-r})^{q+r} \times (x^{r-p})^{r+p}\\[1em] \Rightarrow x^{(p-q)(p+q)} \times x^{(q-r)(q+r)} \times x^{(r-p)(r+p)}\\[1em] \Rightarrow x^{p^2 - q^2} \times x^{q^2 - r^2} \times x^{r^2 - p^2}\\[1em] \Rightarrow x^{(p^2 - q^2) + (q^2 - r^2) + (r^2 - p^2)}\\[1em] \Rightarrow x^{p^2 - q^2 + q^2 - r^2 + r^2 - p^2}\\[1em] \Rightarrow x^{0}\\[1em] \Rightarrow 1$

Hence, $\mathbf{\Big(\dfrac{x^p}{x^q}\Big)^{p+q} \times \Big(\dfrac{x^q}{x^r}\Big)^{q+r} \times \Big(\dfrac{x^r}{x^p}\Big)^{r+p} = 1}$.

Evaluate $\dfrac{x^p}{x^p + x^q} + \dfrac{1}{x^{p-q} + 1}$

Answer

Consider the second term.

Since $x^{p-q} = \dfrac{x^p}{x^q}$,

$\Rightarrow \dfrac{1}{x^{p-q} + 1}\\[1em] \Rightarrow \dfrac{1}{\dfrac{x^p}{x^q} + 1}\\[1em] \Rightarrow \dfrac{1}{\dfrac{x^p + x^q}{x^q}}\\[1em] \Rightarrow \dfrac{x^q}{x^p + x^q}$

Given,

$\Rightarrow \dfrac{x^p}{x^p + x^q} + \dfrac{1}{x^{p-q} + 1}\\[1em] \Rightarrow \dfrac{x^p}{x^p + x^q} + \dfrac{x^q}{x^p + x^q}\\[1em] \Rightarrow \dfrac{x^p + x^q}{x^p + x^q}\\[1em] \Rightarrow 1$

Hence, $\mathbf{\dfrac{x^p}{x^p + x^q} + \dfrac{1}{x^{p-q} + 1} = 1}$.