If $(\frac{8}{3})^{-5} \times (\frac{16}{21})^{5}=(\frac{2}{7})^{x}$, then find $x^{3}$.

Given :

The given expression is $(\frac{8}{3})^{-5} \times (\frac{16}{21})^{5}=(\frac{2}{7})^{x}$.

To do :

We have to find the value of $x^{3}$.

Solution :

$(\frac{8}{3})^{-5} \times (\frac{16}{21})^{5}=(\frac{2}{7})^{x}$

$\Rightarrow (\frac{3}{8})^{5} \times (\frac{16}{21})^{5}=(\frac{2}{7})^{x}$     $[(\frac{a}{b})^{-m} = (\frac{b}{a})^m]$

$\Rightarrow (\frac{3\times 16}{8\times 21})^5=(\frac{2}{7})^{x}$                        $[a^m \times b^m = (ab)^m]$

$\Rightarrow (\frac{2}{7})^{5}=(\frac{2}{7})^{x}$

On comparing, $x=5$.

$x^3 = 5^3 = 125$

Therefore, the value of $x^3$ is 125.

$\left(\frac{8}{3}\right)^{-5} \times \left(\frac{16}{21}\right)^{5} =\left(\frac{2}{7}\right)^{x}$

$$
$$$x^{3}$.

Tutorialspoint

Updated on: 10-Oct-2022

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