Question

Find the value of ${\left(4^{-1}\right)}^4\times 2^5\times {\left(\frac{1}{16}\right)}^3\times {\left(8^{-2}\right)}^5\times {\left({64}^2\right)}^3$

Answer 1

The given expression ${\left(4^{-1}\right)}^4\times 2^5\times {\left(\frac{1}{16}\right)}^3\times {\left(8^{-2}\right)}^5\times {\left({64}^2\right)}^3$ can be simplified as follows:

${\left(4^{-1}\right)}^4\times 2^5\times {\left(\frac{1}{16}\right)}^3\times {\left(8^{-2}\right)}^5\times {\left({64}^2\right)}^3=4^{-1\times 4}\times 2^5\times \frac{1^3}{{16}^3}\times 8^{-2\times 5}\times {64}^{2\times 3}(\because {\left(x^m\right)}^n=x^{m\times n})=4^{-4}\times 2^5\times \frac{1}{{16}^3}\times 8^{-10}\times {64}^6={(2\times 2)}^{-4}\times 2^5\times \frac{1}{(2\times 2\times 2\times 2{)}^3}\times {(2\times 2\times 2)}^{-10}\times {(2\times 2\times 2\times 2\times 2\times 2)}^6$

$=(2^2{)}^{-4}\times 2^5\times \frac{1}{(2^4{)}^3}\times {\left(2^3\right)}^{-10}\times {\left(2^6\right)}^6=2^{-8}\times 2^5\times \frac{1}{2^{12}}\times 2^{-30}\times 2^{36}(\because {\left(x^m\right)}^n=x^{m\times n})=2^{-8}\times 2^5\times 2^{-12}\times 2^{-30}\times 2^{36}(\because \frac{1}{x^n}=x^{-n})=2^{-8+5-12-30+36}(\because x^m\times x^n=x^{m+n})$

$=2^{41-50}=2^{-9}=\frac{1}{2^9}(\because \frac{1}{x^n}=x^{-n})=\frac{1}{512}$

Hence ${\left(4^{-1}\right)}^4\times 2^5\times {\left(\frac{1}{16}\right)}^3\times {\left(8^{-2}\right)}^5\times {\left({64}^2\right)}^3=\frac{1}{512}$