If $\left(\frac{7}{5}\right)^{3} \ \times \ \left(\frac{7}{5}\right)^{x\ +\ 2} \ =\ \left(\frac{7}{5}\right)^{3}$ , then find the value of x.

Given: $\left(\frac{7}{5}\right)^{3} \ \times \ \left(\frac{7}{5}\right)^{x\ +\ 2} \ =\ \left(\frac{7}{5}\right)^{3}$

To find: We have to find the value of x.

Solution: 

$\left(\frac{7}{5}\right)^{3} \ \times \ \left(\frac{7}{5}\right)^{x\ +\ 2} \ =\ \left(\frac{7}{5}\right)^{3}$

$\Longrightarrow \ \left(\frac{7}{5}\right)^{x\ +\ 3\ +\ 2} =\ \left(\frac{7}{5}\right)^{3}$

Since bases are equal we need to compare the powers.

Therefore, $x + 3 + 2 = 3$

=> $x = -2$.

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Updated on: 10-Oct-2022

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