Find the value of m in $$\frac{2}{7}^{-3}×\frac{2}{7}^{-5}=\frac{2}{7}^{-3m+1}$$​​

Correct option is B

​Given:

$$\left( \frac{2}{7} \right)^{-3} \times \left( \frac{2}{7} \right)^{-5} = \left( \frac{2}{7} \right)^{-3m+1}$$​

Formula Used:

$$a^m \times a^n = a^{m+n}$$​

Solution:
Combine exponents on the left-hand side:

​$$\left( \frac{2}{7} \right)^{-3} \times \left( \frac{2}{7} \right)^{-5} = \left( \frac{2}{7} \right)^{-3-5} = \left( \frac{2}{7} \right)^{-8}$$​​

Equating exponents:

​$$-8 = -3m + 1 \\\ -3m = -8 - 1 = -9 \\\ m = \frac{-9}{-3} = 3$$​​