Solve the following

Question:

$\left(\frac{2}{3}\right)^{-5} \times\left(\frac{5}{7}\right)^{-5}$ is equal to

(a) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-10}$

(b) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-5}$

(c) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{25}$

(d) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-25}$

Solution:

(b) $\left(\frac{2}{3} \times \frac{5}{7}\right)^{-5}$

We have:

$\left(\frac{2}{3}\right)^{-5} \times\left(\frac{5}{7}\right)^{-5}=\left(\frac{2}{3} \times \frac{5}{7}\right)^{-5} \quad \cdots\left((a \times b)^{n}=a^{n} \times b^{n}\right)$

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