Question

How do you find the inverse of `f(x)=2-3 log_4(x+1)`?

Answer 1

`f^(-1)(x) = 4^((2-x)/3)-1`

Explanation:

To find the inverse of a function `f(x)` we can let `y = f(x)` and then solve for `x` to obtain `x = f^(-1)(y)`
(To see why the inverse function is obtained, substitute `f(x)` back in for `y` and note that the new function applied to the original returns `x`).

Applying this here:

Let `y = f(x) = 2-3log_4(x+1)`

`=> 3log_4(x+1) = 2 - y`

`=> log_4(x+1) = (2-y)/3`

`=>4^(log_4(x+1)) = 4^((2-y)/3)`

`=> x+1 = 4^((2-y)/3)`

`=> x = 4^((2-y)/3)-1`

Thus we have `f^(-1)(y) = 4^((2-y)/3)-1`

meaning

`f^(-1)(x) = 4^((2-x)/3)-1`

Answer 2

`y=4^((2-x)/3)-1`

Explanation:

Rewrite as `y=2-3log_4(x+1)`.

Swap the `x` and `y`.

`x=2-3log_4(y+1)`

Solve for `y`.

`x-2=-3log_4(y+1)`

`(2-x)/3=log_4(y+1)`

`4^((2-x)/3)=y+1`

`y=4^((2-x)/3)-1`

This can be solved another way. Return to `x-2=-3log_4(y+1)`.

`x-2=log_4(y+1)^-3`

`4^(x-2)=(y+1)^-3`

Raise both sides to the `-1/3` power.

`4^(-1/3(x-2))=y+1`

Again, we get:

`y=4^((2-x)/3)-1`