Aus Online Mathematik Brückenkurs 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 numbers,

\displaystyle \ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)

Using the log law, 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\centerdot \ln a, we can get rid of \displaystyle x from the exponents:

\displaystyle \ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3

Collecting together \displaystyle x on one side and the other terms on the other,

\displaystyle x\ln e+x\ln 3=\ln 2-\ln 13

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

\displaystyle x\left( 1+\ln 3 \right)=\ln 2-\ln 13

Then, solve for \displaystyle x

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

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

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

in order to indicate that \displaystyle x is negative.