Home > Precalculus > Logarithm-- Inverse of an Exponential Function

How do you find the inverse of #f(x) = 3log(x-1)#?

Answer 1

I found: #f(x)=1+10^(x/3)#

I am not sure it is the "formal" way to do it but I do it like this: I try to "extract" #x#: #log(x-1)=(f(x))/3# assuming base #10# for the log, I can write: #x-1=10^(f(x)/3)# #x=1+10^(f(x)/3)# now I switch #x# and #f(x)# to get: #f(x)=1+10^(x/3)# Graphically: #f(x)=3log(x-1)# graph{3log(x-1) [-10, 10, -5, 5]} and: #f(x)=1+10^(x/3)# graph{1+10^(x/3) [-10, 10, -5, 5]}

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Answer 2

Elijah Allen

To find the inverse of (f(x) = 3\log(x - 1)), follow these steps:

Replace (f(x)) with (y).
Swap (x) and (y) to interchange the dependent and independent variables.
Solve the resulting equation for (y).
Replace (y) with (f^{-1}(x)) to express the inverse function.

Here's how it's done:

Start with (y = 3\log(x - 1)).
Swap (x) and (y) to get (x = 3\log(y - 1)).
Solve for (y):

[x = 3\log(y - 1)]

[x/3 = \log(y - 1)]

[10^{x/3} = y - 1]

[y = 10^{x/3} + 1]

Replace (y) with (f^{-1}(x)) to express the inverse function:

[f^{-1}(x) = 10^{x/3} + 1]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.