Question

Given the function f(x) = log3(x + 4), find the value of f−1(3

Answer 1

First of all, find f ⁻¹ (x), or the inverse function like so:
y = log(3x + 12)
e^y = 3x + 12
3x = e^y - 12
x = 1/3 (e^y - 12)
Now, swap x and y:
f ⁻¹ (x) = y = 1/3 (e^x - 12)

Secondly and finally, substitute 3 into this eq'n:
f ⁻¹ (3) = 1/3 (e³ -12) = 2.695.....
so 2.695 (to 4 sig.figs/ 3 d.ps)
or if asked for an exact answer then it is just: 1/3 (e³ -12), which can be simplfied to 1/3e³ - 4.

Answer 2

23

Explanation:

Given :
`f(x) = log_3(x + 4)`

To Find:
`f^(-1)(3)`

Solution:

`y = log_3(x + 4)`

Replace y with x and x with y

`x = log_3(y + 4)`

`3^x = y + 4`

`3^x-4 = y`

So,
`f^(-1)(x)=3^x-4`

Now substitute x= 3

`f^(-1)(3)=3^(3)-4`

`f^(-1)(3)=27-4`

`f^(-1)(3)=23`

Hence the value of
`f^(-1)(3)` is 23