Question

Solve the following equation:
${\left(\frac{4}{9}\right)}^x\cdot {\left(\frac{27}{8}\right)}^{x-1}=\frac{\log 4}{\log 8}$

Answer 1

$\left(\frac{2}{3}{)}^{2x}\right(\frac{3}{2}{)}^{3x-3}=\frac{2log2}{3log2}$

$(\frac{2}{3}{)}^{3-x}=2/3$

$3-x=1$

$x=2$