Question

did I solve this logarithmic equation right? If not, tell me how to do so. It asks to solve the answer in terms of x

Answer 1

Given:

`\log _(4^x)2^a=3`

To solve this, we will apply several laws of logarithm along the line.

`\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}`
`\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=(3x\log 4)/(\log 2) \end{gathered}`

Also applying another law of logarithm below;

`\frac{\text{logb}}{\log a}=\log _ab`
`\begin{gathered} \text{Hence,} \\ a=(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}`

Therefore, the solution in terms of x is;

`a=6x`