Given a logarithmic function in the form
[tex]\log ^{}_{B^{}}A^x[/tex]
It is simplified according to the power law of logarithm as
[tex]x\times\log _BA=x\log _BA[/tex]
When it is in the form
[tex]\log ^{}_{B^y}A^{}[/tex]
It is simplified according to the Base power law of logarithm as
[tex]\frac{1}{y}\times\log _BA=\frac{1}{y}\log _BA[/tex]
Thus,
[tex]\log _{4^x}2^a=3[/tex]
...is simplified as
[tex]\begin{gathered} a\times\frac{1}{x}\times\log _42=3 \\ \frac{a}{x}\times\log _42=3 \\ This\text{ is further simplified as} \\ \frac{a}{x}\times\log _{2^2}2=3 \\ \frac{a}{x}\times\frac{1}{2}\times\log _22=3 \\ \\ \\ \end{gathered}[/tex]
But
[tex]\log _AA=1[/tex]
Hence, we have
[tex]\begin{gathered} \frac{a}{x}\times\frac{1}{2}\times1=3 \\ \frac{a}{2x}=3 \\ a=6x \end{gathered}[/tex]
Hence the solution in terms of x is given as a=6x
