1. ## Inverse function

Hi, I could really use some help with this. I couldn't understand the book's way of solving the problem.

Find the inverse function of the following, also state the domain and range of the original functions:

a) g(x)=3+e^x
b) h(x)=log(x-1)

2. ## Re: Inverse function

Originally Posted by FatimaA
Hi, I could really use some help with this. I couldn't understand the book's way of solving the problem.

Find the inverse function of the following, also state the domain and range of the original functions:

a) g(x)=3+e^x
b) h(x)=log(x-1)

Are you familiar with the general procedure for finding inverse functions? Interchange the $x$ and the $y$ and solve for $y$. This may involve taking the logarithm of both sides, or exponentiating both sides.

So for (a), we start with $y = 3+e^x$ and we switch the $x$'s and $y$'s, producing

$x = 3+e^y$.

Now, can you solve the above for $y$?

3. ## Re: Inverse function

Yeah I'm familiar with interchanging the x and y but I'm pretty sure my teacher wants us to take the logarithm of both sides, which I am really, really confused about. I don't know what you mean by exponentiating.

For y= 3+e^x I thought you were supposed to get the x on one side and then switch the variables?

4. ## Re: Inverse function

Originally Posted by FatimaA
For y= 3+e^x I thought you were supposed to get the x on one side and then switch the variables?
Yes, you could solve for $x$ and then switch the variables after; that accomplishes the same thing. So the problem is how to solve the equation $y = 3+e^x$ for $x$. Begin by isolating the exponential term:

$y=3+e^x\quad\Rightarrow\quad e^x=y-3$

Now take the natural logarithm of both sides:

$e^x=y-3\quad\Rightarrow\quad\ln e^x=\ln(y-3)$

And $\ln e^x$ is just $x$ (i.e., these are inverse operations). So we are left with

$x = \ln(y-3)$.

For the second one, I assume that the base is 10. You can cancel out the operation of the logarithm by raising 10 to the power of each side:

$y=\log(x-1)\quad\Rightarrow\quad10^y=10^{\log(x-1)}\quad\Rightarrow\quad10^y=x-1$.

Basically, you are converting the equation from logarithmic form $\log_bx=a$ to the exponential form $b^a=x$.

5. ## Re: Inverse function

Thanks! Actually I found the procedure in my notes lol it was there all along but now I can check my work. I did exactly what you did, but this way:

1) y=3+e^x

2) y-3=e^x

3) ln(y-3)=ln(e^x)

4) x=ln(y-3)

5) g^-1(x)=ln(x-3)

I got the same answer as you for the second one as well. I did it pretty much the same way:

1) y=log(x-1)

2) 10^y=10^log(x-1)

3) 10^y=x-1

4) (10^y)+1=x

5) h^-1(x)=(10^x)+1

Obviously I took many more steps but it looks like we both got the same thing.