From Förberedande kurs i matematik 1

In the equation, both sides are positive because the factors \displaystyle e^{x} and \displaystyle 3^{-x} are positive regardless of the value of \displaystyle x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,

\displaystyle \ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}

Using the log laws, we can divide up the products into several logarithmic terms,

\displaystyle \ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},

and using the law \displaystyle \ln a^{b}=b\cdot \ln a, we can get rid of \displaystyle x from the exponents

\displaystyle \ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}

Collect \displaystyle x on one side and the other terms on the other,

\displaystyle x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}

Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1,

\displaystyle x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}

Then, solve for \displaystyle x,

\displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}

Note: Because \displaystyle \ln 2 < \ln 13, we can write the answer as

\displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3}

to indicate that \displaystyle x is negative.