Question

Find m.

$\frac{{\left(3\right)}^{2m+1}}{27}\left(9^2\right)={\left(\left({-3}^8\right)3^4\right)}^5$ ?

Answer 1

$\left(\frac{\left(3{)}^{2m+1}\right)}{27}\right)\times 9^2=\left(\right(-3{)}^8{3}^4{)}^5\left(\frac{3^{2m+1}}{3^3}\right)\times 3^4=(3^{12}{)}^5{3}^{2m+1+4-3}=3^{60}2m+2=60\therefore m=29$