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

Answer:

23

Step-by-step 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