Answer:
a=6x
Explanation:
Given the logarithmic equation:
[tex]\log _{4^x}2^a=3[/tex]
We appy the change of base rule:
[tex]\begin{gathered} \log _ab=\frac{\log b}{\log a} \\ \implies\log _{4^x}2^a=\frac{\log2^a}{\log4^x}=3 \end{gathered}[/tex]
Next, rewrite 4 as a power of 2.
[tex]\begin{gathered} \frac{\log2^a}{\log4^x}=3 \\ \implies\frac{\log 2^a}{\log 2^{2x}}=3 \end{gathered}[/tex]
Take the index of the numbers as the product by the index law.
[tex]\frac{a\log 2}{2x\log 2}=3[/tex]
Cancel out log 2 in the numerator and denominator
[tex]\implies\frac{a}{2x}=3[/tex]
Finally, cross multiply to express "a" in terms of x.
[tex]\begin{gathered} a=3\times2x \\ a=6x \end{gathered}[/tex]
