Question

Solve:-

$(\frac{2^5}{2^8})\times 2^{-7}$

• A. $2^{10}$
• B. $2^6$
• C. $2^{-10}$
• D. $2^{-4}$

Answer 1

The correct option is C

$2^{-10}$

$\frac{\left(2^5\right)}{\left(2^8\right)}\times \left(2^{-7}\right)$

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

$=\frac{\left(2^{5+(-7)}\right)}{\left(2^8\right)}$ $[\because a^m\times a^n=a^{m+n}]$

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

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

$=2^{(-2-8)}$ $[\because a^m÷a^n=a^{m-n}]$

$=2^{-10}$