Given the expression:
[tex]\ln 3+\ln 9[/tex]the given logarithms are written to the base (e), we will change the base to 3
First, we will simplify the expression then change the base to 3
As we know: ln(AB) = ln A + ln B
so, the given expression can be written as:
[tex]\ln 3+\ln 9=\ln (3\cdot9)=\ln 27[/tex]now, we will change (ln 27) from the base (e) to the base (3) as follows:
[tex]\begin{gathered} y=\ln 27 \\ e^y=27 \end{gathered}[/tex]Now, taking the logarithm to the base 3
so,
[tex]\begin{gathered} \log _3e^y=\log _327 \\ y\cdot\log _3e=\log _33^3 \\ y\cdot\log _3e=3\cdot\log _33 \\ y\cdot\log _3e=3\cdot1 \\ y\cdot\log _3e=3 \\ \\ y=\frac{3}{\log _3e} \end{gathered}[/tex]so, the answer will be:
[tex]\frac{3}{\log _3e}[/tex]