Question

Please answer me this: Am I correct? In the picture provided, I solved a logarithmic equation and I answered it in terms of x.

Answer 1

a = 6x

Step-by-step explanation:

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

using logarithm rule:

`\begin{gathered} \log _ba\text{ = }\frac{\log a}{\log \text{ b}} \\ \text{applying same to the log we have:} \\ \log \text{ of the number/log of the base} \\ \log _(4^x)2^a\text{ = }(\log2^a)/(\log4^x) \end{gathered}`
`\begin{gathered} \sin ce\text{ }\log _(4^x)2^a=3\text{ } \\ (\log2^a)/(\log4^x)\text{ = 3} \\ \text{Another logarithm rule:} \\ \log a^b\text{ = bloga} \\ \text{Exponent multiply the log:} \\ \text{Applying that to what we have:} \\ \frac{a\log2^{}}{x\log4^{}}\text{ = 3} \end{gathered}`
`\begin{gathered} \text{cross multiply:} \\ a\log 2\text{ = 3(x log4)} \\ \text{alog 2 = 3x log4} \\ \text{collect like terms by first dividing both sides by 3x:} \\ (a\log2)/(3x)\text{ = log 4} \\ \text{Then divide both sides by log2:} \\ (a\log2)/(3x(\log2))=\text{ }(\log4)/(\log2) \\ (a)/(3x)\text{ = }(\log4)/(\log2) \end{gathered}`
`\begin{gathered} \text{SInce 4 = 2}^2, \\ \log 4=log2^2 \\ (a)/(3x)\text{ = }\frac{\log 2^2}{\log \text{ 2}} \\ we\text{ apply the log rule we used before:} \\ (a)/(3x)\text{ = }\frac{2\log 2^{}}{\log \text{ 2}} \\ \log \text{ 2 cancels out:} \\ (a)/(3x)\text{ = }2(1) \\ (a)/(3x)=\text{ 2} \\ a\text{ = 2(3x)} \\ a\text{ = 6x} \end{gathered}`