Answer
$\text{Exact: } \dfrac{ \ln \left(\dfrac{4}{3} \right)}{\ln 5 +\ln \left(\dfrac{4}{3} \right)}$
$\text{Approximately: } 0.152$
Work Step by Step
Take the natural logarithm of both sides:
$\ln \left(\dfrac{4}{3} \right)^{1-x} = \ln 5^{x}$
Since $\ln M^r = r \ln M$, then:
$\therefore\ln \left(\dfrac{4}{3} \right)^{1-x} = (1-x) \ln \left(\dfrac{4}{3} \right) \hspace{20pt} \text{and} \hspace{20pt} \ln 5^{x} = x \ln 5$
Thus, the equation above is equivalent to :
$(1-x) \ln \left(\dfrac{4}{3} \right) =x \ln 5$
$\ln \left(\dfrac{4}{3} \right) -x \ln \left(\dfrac{4}{3} \right) = x \ln 5$
Combine the $x$ terms:
$\ln \left(\dfrac{4}{3} \right) = x \ln 5 +x \ln \left(\dfrac{4}{3} \right)$
Factor out $x$ (the common factor), then solve for $x$:
$\ln \left(\dfrac{4}{3} \right) = x \left(\ln 5 +\ln \left(\dfrac{4}{3} \right) \right)$
$x = \boxed{\dfrac{ \ln \left(\dfrac{4}{3} \right)}{\ln 5 +\ln \left(\dfrac{4}{3} \right)} \approx 0.152}$
