Find the value of $x$:

$\left(\frac{-8}{9}\right)^{15}\ \div\ \left(\frac{-8}{9}\right)^{x}\ =\ \left(\frac{-8}{9}\right)^{2}$

Given: $\left(\frac{-8}{9}\right)^{15}\ \div\ \left(\frac{-8}{9}\right)^{x}\ =\ \left(\frac{-8}{9}\right)^{2}$

To find: Here we have to find the value of $x$.

Solution:

$\left(\frac{-8}{9}\right)^{15} \ \div \ \left(\frac{-8}{9}\right)^{x} \ =\ \left(\frac{-8}{9}\right)^{2}$

Using the division rule of exponent [$a^x\ \div\ a^y$ = $a^{x\ -\ y}$]  

$\left(\frac{-8}{9}\right)^{15\ -\ x}  =\left(\frac{-8}{9}\right)^{2}$

Since the base is same we can compare the power. So,

$15\ -\ x\ =\ 2$

$x\ =\ 15\ -\ 2$

$x\ =$ 13

So, value of $x$ is 13.

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Updated on: 10-Oct-2022

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