Home > Precalculus > Function Composition

How do you find the inverse of #f(x) = 2log (3x-12) + 5#?

Answer 1

Serenity Johnson

To find the inverse of the function f(x) = 2log(3x - 12) + 5, follow these steps:

Replace f(x) with y: y = 2log(3x - 12) + 5
Swap x and y: x = 2log(3y - 12) + 5
Solve the equation for y: x - 5 = 2log(3y - 12)
Divide both sides by 2: (x - 5) / 2 = log(3y - 12)
Rewrite the equation in exponential form: 10^((x - 5) / 2) = 3y - 12
Add 12 to both sides: 10^((x - 5) / 2) + 12 = 3y
Divide both sides by 3: [10^((x - 5) / 2) + 12] / 3 = y
Replace y with f^(-1)(x): f^(-1)(x) = [10^((x - 5) / 2) + 12] / 3

Therefore, the inverse of the function f(x) = 2log(3x - 12) + 5 is f^(-1)(x) = [10^((x - 5) / 2) + 12] / 3.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Allen

The inverse function #f^-1(x)# is #10^((x-5)/2)/3+4#.

If you let #y=2log(3x-12)+5#, make a new equation switching the #x#'s and #y#'s and then solve for the new #y#:

#x=2log(3y-12)+5#

#x-5=2log(3y-12)#

#(x-5)/2=log(3y-12)#

Convert to exponential form:

#10^((x-5)/2)=3y-12#

#10^((x-5)/2)+12=3y#

#10^((x-5)/2)/3+4=y#

That is the inverse function. Hope this helped!

By signing up, you agree to our Terms of Service and Privacy Policy

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.