Exponents #6

$\Rightarrow \left(\dfrac{4}{9}\right)^x \times \left(\dfrac{27}{8}\right)^{x-1} = \dfrac{\log 4}{\log 8}$ $\Rightarrow \left(\dfrac{2}{3}\right)^{2x} \times \left(\dfrac{3}{2}\right)^{3(x-1)} = \dfrac{\log 2^2}{\log 2^3}$ $\Rightarrow \left(\dfrac{2}{3}\right)^{2x} \times \left(\dfrac{2}{3}\right)^{3-3x} = \dfrac{2}{3} \dfrac{\log 2}{\log 2}$ $\Rightarrow \left(\dfrac{2}{3}\right)^{2x + 3 - 3x} = \left(\dfrac{2}{3}\right)^1$ Bases are same, equating powers.
$\Rightarrow 3 - x = 1 \Rightarrow \boxed{ x = 2 }$

${ \left( \frac { 4 }{ 9 } \right) }^{ x }\times { \left( \frac { 27 }{ 8 } \right) }^{ x-1 }=\frac { \log { 4 } }{ \log { 8 } } \\ \\ \quad { \left( \frac { 4 }{ 9 } \right) }^{ x }\times { \left( \frac { 27 }{ 8 } \right) }^{ x-1 }=\frac { 2 }{ 3 } \\ \\ { \left( \frac { 2 }{ 3 } \right) }^{ 2x }\times { \left( \frac { 2 }{ 3 } \right) }^{ -3s+3 }=\frac { 2 }{ 3 } \\ \\ \quad \quad \quad \quad \quad \Leftrightarrow 2x-3x+3=1\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \boxed { x=2 }$