Question

Write in single exponent form:

${(-2)}^5\times b^5$

Answer 1

In ${(-2)}^5\times b^5$, there are different bases but the exponents are same.

So, for any non-zero integers $a,b$ where $m$ is any whole number,

$a^m\times b^m={\left(ab\right)}^m$

${(-2)}^5\times b^5$$=\left[\right(-2)\times (-2)\times (-2)\times (-2)\times (-2\left)\right]\times (b\times b\times b\times b\times b)$ (Here $a=-2,b=b,m=5$)

$=\left[\right(-2)\times b]\times \left[\right(-2)\times b]\times \left[\right(-2)\times b]\times \left[\right(-2)\times b]\times \left[\right(-2)\times b]$

$={\left(-2\times b\right)}^5$

$={(-2b)}^5$

Therefore, ${(-2)}^5\times b^5$$={(-2b)}^5$.