Question

Fill in the blank to make the statement true:

${\left(\frac{-1}{4}\right)}^3\times {\left(\frac{-1}{4}\right)}^{\_\_}={\left(\frac{-1}{4}\right)}^{11}$

Answer 1

Find the missing exponent.

Given equation: ${\left(\frac{-1}{4}\right)}^3\times {\left(\frac{-1}{4}\right)}^{\_\_}={\left(\frac{-1}{4}\right)}^{11}$

We know that for any non-zero integer $‘a’$ and whole numbers $m$ and $n$, $a^m\times a^n=a^{m+n}$

By comparing the given equation with $a^m\times a^n=a^{m+n}$, we can say $a=\frac{-1}{4},m=3$ and $m+n=11$

By substituting the value of $m$ in $m+n=11$ we get,

$$\begin{gathered} 3+n=11 \\ \Rightarrow n=11-3 \\ \Rightarrow n=8 \end{gathered}$$

So, the missing exponent is $8$.

Hence, ${\left(\frac{-1}{4}\right)}^3\times {\left(\frac{-1}{4}\right)}^{\underline{8}}={\left(\frac{-1}{4}\right)}^{11}$