Question
Simplify: $(2^5÷ 2^8) × 2^{-7}$

Answer

Using laws of exponents, $a^m \div a^n=(a)^{m-n}$ and $a^m \times a^n=(a)^{m+n}$[$\because$ a is non-zero integer] $\therefore$ $(2^5\div2^8)\times(-2)^{-7}=(2)^{5-8}\times(2)^{-7}$
$=(2)^{-3}\times(2)^{-7}$ $=(2)^{-3-7}=(2)^{-10}$
$=\frac{1}{2^{10}}=\frac{1}{1024}$
$\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}}^\text{m}\Big]$

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