Question
Find x.$-\frac{1}{7}^{-5}+-\frac{1}{7}^{-7}=(-7)^{\text{x}}$
✓
Answer
Using law of exponents, $ a^m \times a^n=a^{m+n}[\because$ a is non-zero integer$]$
Then, $\Big(-\frac{1}{7}\Big)^{-5+7}=(-7)^{\text{x}}$
$\Rightarrow \ \big(\frac{-1}{7}\big)^2=(-7)^{\text{x}}$
$\Rightarrow \ (-7\big)^2=(-7)^{\text{x}}$
On comparing powers of $(-7)$, we get $x = -2$
Then, $\Big(-\frac{1}{7}\Big)^{-5+7}=(-7)^{\text{x}}$
$\Rightarrow \ \big(\frac{-1}{7}\big)^2=(-7)^{\text{x}}$
$\Rightarrow \ (-7\big)^2=(-7)^{\text{x}}$
On comparing powers of $(-7)$, we get $x = -2$
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