[tex]a=\log _24^{3x}[/tex]
1) In this logarithm we can notice there are two variables. Let's rewrite it by applying the definition for Logarithm:
[tex]\begin{gathered} \log _{4^x}2^a=3 \\ 2^a=(4^x)^3 \end{gathered}[/tex]
2) Now to solve for a, we need to perform some logarithm manipulation:
[tex]\begin{gathered} \log _22^a=\log _24^{3x} \\ a\cdot\log _22=\log _24^{3x} \\ a\cdot1=\log _24^{3x} \\ a=\log _24^{3x} \end{gathered}[/tex]
Note here, the application of logarithm of a power "dropping" the exponent in front of the logarithm expression and rewriting it as a factor.
3) Hence, the answer is:
[tex]a=\log _24^{3x}[/tex]
