Question

I need help on thisI need it answered in terms of x,Step by step

Answer 1

Given a logarithmic function in the form

`\log ^{}_{B^{}}A^x`

It is simplified according to the power law of logarithm as

`x*\log _BA=x\log _BA`

When it is in the form

`\log ^{}_(B^y)A^{}`

It is simplified according to the Base power law of logarithm as

`(1)/(y)*\log _BA=(1)/(y)\log _BA`

Thus,

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

...is simplified as

`\begin{gathered} a*(1)/(x)*\log _42=3 \\ (a)/(x)*\log _42=3 \\ This\text{ is further simplified as} \\ (a)/(x)*\log _(2^2)2=3 \\ (a)/(x)*(1)/(2)*\log _22=3 \\ \\ \\ \end{gathered}`

But

`\log _AA=1`

Hence, we have

`\begin{gathered} (a)/(x)*(1)/(2)*1=3 \\ (a)/(2x)=3 \\ a=6x \end{gathered}`

Hence the solution in terms of x is given as a=6x