Exponents and Powers — MATHS STD 7 — Question

Solution:

Here, $\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}$

$=\Big(\frac{6}{13}\Big)^{0}+\Big(\frac{6}{13}\Big)^{5\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}+\text{a}^{\text{mn}}$

$\Big(\frac{6}{13}^{10}+\Big(\frac{6}{13}\Big)^{10}=\Big(\frac{6}{13}\Big)^{10-10}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}},(\text{m}>\text{n})\big]$

$=\Big(\frac{6}{13}\Big)^{0}$

$\therefore\Big(\frac{6}{13}\Big)^{10}+\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$