Question

(3/4)^2m-5 = (27/64)^2-3m

Answer 1

$\tt\color{green}{Given }$

$\Big( \frac{3}{4}\Big)^{2m-5} = \Big( \frac{27}{64}\Big)^{2-3m}$

$\implies \Big( \frac{3}{4}\Big)^{2m-5} = \Big( \frac{3^{3}}{4^{3}}\Big)^{2-3m}$

$\implies \Big( \frac{3}{4}\Big)^{2m-5} = \Big( \frac{3}{4}\Big)^{3(2-3m)}$

$\boxed{ \pink{ \because \frac{x^{m}}{y^{m}} = \big( \frac{x}{y}\big)^{m} }}$

$\implies \Big( \frac{3}{4}\Big)^{2m-5} = \Big( \frac{3}{4}\Big)^{6 -9m)}$

$\implies 2m - 5 = 6 - 9m$

$\boxed{\blue{ If \: a^{m} = a^{m} \implies m = n }}$

$\implies 2m + 9m = 6 + 5$

$\implies 11m = 11$

$\implies m = \frac{11}{11}$

$\implies m= 1$

$\tt\color{blue}{Therefore.,}$

$\sf\color{red}{ Value \: of\: m}\green { = 1 }$

Thankyou :)