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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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