Aus Online Mathematik Brückenkurs 1

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

First, we take logs of both sides,

\displaystyle \ln \left( 3e^{x} \right)=\ln \left( 7\centerdot 2^{x} \right)

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

\displaystyle \ln 3+x\centerdot \ln e=\ln 7+x\centerdot \ln 2

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

\displaystyle x\left( \ln e-\ln 2 \right)=\ln 7-\ln 3

The solution is now

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