Solution:
Given:
[tex]\log _{4^x}2^a=3[/tex]
To solve this, we will apply several laws of logarithm along the line.
[tex]\begin{gathered} \log _{4^x}2^a=3 \\ \text{Apply ing the law; if} \\ \log _ab=x,\text{ then} \\ a^x=b \\ \text{Hence,} \\ (4^x)^3=2^a \\ 4^{3x}^{}=2^a \end{gathered}[/tex][tex]\begin{gathered} \text{Taking the log of both sides,} \\ \log 4^{3x}=\log 2^a \\ \\ \text{Apply ing the law;} \\ \log a^x=x\log a \\ \text{Then the equation becomes;} \\ 3x\log 4=a\log 2 \\ \text{Dividing both sides by log2,} \\ a=\frac{3x\log 4}{\log 2} \end{gathered}[/tex]
Also applying another law of logarithm below;
[tex]\frac{\text{logb}}{\log a}=\log _ab[/tex][tex]\begin{gathered} \text{Hence,} \\ a=\frac{3x\log 4}{\log 2} \\ a=3x(\log _24) \\ a=3x(\log _22^2) \\ Thus,\text{ it becomes} \\ a=3x.2\log _22 \\ a=6x\log _22 \\ \\ \text{Apply ing the rule;} \\ \log _aa=1 \\ \log _22=1 \\ \text{Hence, } \\ a=6x\log _22 \\ a=6x.1 \\ a=6x \end{gathered}[/tex]
Therefore, the solution in terms of x is;
[tex]a=6x[/tex]
