Answer
$\text{Exact: } \dfrac{ \ln 5 - \ln 2}{\ln 2 + 2 \ln 5}$
$\text{Approximately: } 0.234$
Work Step by Step
Take the natural logarithm of both sides:
$\ln 2^{x+1} = \ln 5^{1-2x}$
Since $\ln M^r = r \ln M$, then:
$\ln 2^{x+1} = (x+1) \ln 2 \hspace{20pt} \text{and} \hspace{20pt} \ln 5^{1-2x} = (1-2x) \ln 5$
Thus the equation above is equivalent to
$(x+1) \ln 2 =(1-2x) \ln 5$
Distribute to obtain:
$x \ln 2 + \ln 2 = \ln 5 -2x \ln 5$
Combine the $x$ terms:
$x \ln 2+ 2x \ln 5 = \ln 5 - \ln 2$
Factor out the $x$ (the common factor) then solve for $x$:
$x(\ln 2 + 2 \ln 5 ) = \ln 5 - \ln 2\\$
$x = \boxed{\dfrac{ \ln 5 - \ln 2}{\ln 2 + 2 \ln 5} \approx 0.234}$
