Question

How do you find the inverse of `y=ln(8x + 1)`?

Answer 1

`x=(e^y-1)/8`

Explanation:

If `y=ln(8x+1)`
then (by definition)
`color(White)("XXX")e^y=8x+1`

From this it follows that
`color(White)("XXX")e^y-1 = 8x`
and
`color(White)("XXX")x=(e^y-1)/8`

It is important to remember:
`color(White)("XXX")log_b a = c <=> b^c = a`

(and that `ln x = log_e x`)