Question

How do I solve this? y = 4 Log x + 5

Answer 1

`\\ \tt\hookrightarrow 4logx+5=y`

`\\ \tt\hookrightarrow 4logx=y-5`

`\\ \tt\hookrightarrow logx=(y-5)/(4)`

`\\ \tt\hookrightarrow x=10^{(y-5)/(4)}`

`\\ \tt\hookrightarrow x=\sqrt[4]{10^(y-5)}`

Answer 2

`x=10^(\frac14(y-5))`

Explanation:

`y = 4 log (x) + 5`

`\implies y - 5 = 4 log(x)`

`\implies \frac14(y-5)=log(x)`

Using log law
`log_a(b)=c \implies a^c=b`:

`x=10^(\frac14(y-5))`