MCQ
$\Big(-\frac{5}{7}\Big)^{-5}$ is equal to:
- A$\Big(-\frac{5}{7}\Big)^{-5}$
- B$\Big(\frac{5}{7}\Big)^5$
- C$\Big(\frac{7}{5}\Big)^5$
- ✓$\Big(\frac{-7}{5}\Big)^5$
✓
Answer
Correct option: D.
$\Big(\frac{-7}{5}\Big)^5$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}} [\because$ a is non$-$zero integer$]$
$\therefore$ $\Big(-\frac{5}{7}\Big)^5=\frac{1}{\Big(\frac{-5}{7}\Big)^5}$
$=\Big(-\frac{7}{5}\Big)^5$
$\therefore$ $\Big(-\frac{5}{7}\Big)^5=\frac{1}{\Big(\frac{-5}{7}\Big)^5}$
$=\Big(-\frac{7}{5}\Big)^5$
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