Given:
[tex]ln\mleft(e^{2x}\mright)+ln\mleft(e^x\mright)[/tex]
To simplify:
Applying the log rule,
[tex]\log _c\mleft(a\mright)+\log _c\mleft(b\mright)=\log _c\mleft(ab\mright)[/tex]
We get,
[tex]\begin{gathered} ln(e^{2x})+ln(e^x)=\ln (e^{2x}\cdot e^x) \\ =\ln (e^{3x}) \\ =3x(\ln e) \\ =3x(1) \\ =3x \end{gathered}[/tex]
Hence, the answer is 3x.
