Question
$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)$
✓
Answer
$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$Solution:
Here, $\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}$
$=\Big(\frac{6}{13}\Big)^{0}+\Big(\frac{6}{13}\Big)^{5\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}+\text{a}^{\text{mn}}$
$\Big(\frac{6}{13}^{10}+\Big(\frac{6}{13}\Big)^{10}=\Big(\frac{6}{13}\Big)^{10-10}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}},(\text{m}>\text{n})\big]$
$=\Big(\frac{6}{13}\Big)^{0}$
$\therefore\Big(\frac{6}{13}\Big)^{10}+\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$
Here, $\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}$
$=\Big(\frac{6}{13}\Big)^{0}+\Big(\frac{6}{13}\Big)^{5\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}+\text{a}^{\text{mn}}$
$\Big(\frac{6}{13}^{10}+\Big(\frac{6}{13}\Big)^{10}=\Big(\frac{6}{13}\Big)^{10-10}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}},(\text{m}>\text{n})\big]$
$=\Big(\frac{6}{13}\Big)^{0}$
$\therefore\Big(\frac{6}{13}\Big)^{10}+\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$
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