Answer
$\displaystyle \{\frac{\log 4-\log 3}{\log 5+\log 4-\log 3} \}$ or $\{0.152\}$
Work Step by Step
... Apply $\log$(...) to both sides and isolate x
$\displaystyle \log\left(\frac{4}{3}\right)^{1-x}=\log\left(5^{x}\right)$
$(1-x)\displaystyle \log(\frac{4}{3})=x\log 5$
$\displaystyle \log(\frac{4}{3})-x\log(\frac{4}{3})=x\log 5$
$\displaystyle \log(\frac{4}{3})=x\log 5+x\log(\frac{4}{3})$
$\log 4-\log 3=x(\log 5+\log 4-\log 3)$
$x=\displaystyle \frac{\log 4-\log 3}{\log 5+\log 4-\log 3}\approx 0.152$
Solution set: $\displaystyle \{\frac{\log 4-\log 3}{\log 5+\log 4-\log 3} \}$ or $\{0.152\}$
