Question

Find the value of p for which $(\frac{2}{7}{)}^5\times (\frac{2}{7}{)}^{-6}=(\frac{2}{7}{)}^{2p-1}$

Answer 1

$\frac{2^5}{7^5}\times \frac{2^{-6}}{7^{-6}}=(\frac{2}{7}{)}^{2p-1}\frac{2^{5-6}}{7^{5-6}}=(\frac{2}{7}{)}^{2p-1}\frac{2^{-1}}{7^{7^{-1}}}=\left(\frac{2}{7}{)}^{2p-1}\right(\frac{2}{7}{)}^{-1}=(\frac{2}{7}{)}^{2p-1}$
On comparing
$-1=2p–10=2pp=0$