Aus Online Mathematik Brückenkurs 1

The equation has the same form as the equation in exercise b and we can therefore use the same strategy.

First, we take logs of both sides,

\displaystyle \ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}

and use the log laws to make \displaystyle x more accessible,

\displaystyle \ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}

Then, collect together the \displaystyle x terms on the left-hand side,

\displaystyle x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}

The solution is now

\displaystyle x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}