Question

What is the inverse of #y = log_8 (4x) # ?

Answer 1

`f^-1(x)=1/4(8^x)`

Explanation:

`"rearrange making x the subject"`

`[log_bx=nhArrx=b^n]`

`4x=8^y`

`x=1/4xx8^y`

`rArrf^-1(x)=1/4(8^x)`