Question

Fill in the blank to make the statement true:

${\left(\frac{13}{14}\right)}^5÷{\left(\_\_\_\right)}^2={\left(\frac{13}{14}\right)}^3$

Answer 1

Find the missing number.

Given equation: ${\left(\frac{13}{14}\right)}^5÷{\left(\_\_\_\right)}^2={\left(\frac{13}{14}\right)}^3$

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

Let the missing number be $x$.

So the given equation becomes,

$$\begin{gathered} {\left(\frac{13}{14}\right)}^5÷{\left(x\right)}^2={\left(\frac{13}{14}\right)}^3 \\ \Rightarrow \frac{{\left(\frac{13}{14}\right)}^5}{{\left(x\right)}^2}={\left(\frac{13}{14}\right)}^3 \\ \Rightarrow \frac{{\left(\frac{13}{14}\right)}^5}{{\left(\frac{13}{14}\right)}^3}={\left(x\right)}^2\mathbf{} \\ \Rightarrow {\left(\frac{13}{14}\right)}^{5-3}={\left(x\right)}^2\mathbf{[}\mathbf{\because }{\mathit{a}}^{\mathbf{m}}\mathbf{}\mathbf{÷}\mathbf{}{\mathit{a}}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{m}\mathbf{-}\mathbf{n}}\mathbf{]} \\ \Rightarrow {\left(\frac{13}{14}\right)}^2={\left(x\right)}^2 \\ \Rightarrow \frac{13}{14}=x \\ \therefore x=\frac{13}{14} \end{gathered}$$

Hence, ${\left(\frac{13}{14}\right)}^5÷{\left(\underline{\frac{13}{14}}\right)}^2={\left(\frac{13}{14}\right)}^3$