1. ## Inverse Exponential Function?

Ive been having a bit of trouble with this one.

Find the Inverse.

$\displaystyle g(x)= 3^x + 1$

I think that the answer is this but Im not sure

$\displaystyle g^-1(x)=\log_{x}3 + 1$

$\displaystyle g^-1(x)=\log_{x}3 - 1$

Thanks for any and all help given.

2. Originally Posted by Legal Eagle
Ive been having a bit of trouble with this one.

Find the Inverse.

$\displaystyle g(x)= 3^x + 1$

I think that the answer is this but Im not sure

$\displaystyle g^-1(x)=\log_{x}3 + 1$

$\displaystyle g^-1(x)=\log_{x}3 - 1$

Thanks for any and all help given.
neither is correct

$\displaystyle y = 3^x + 1$

for inverse, switch x and y and solve for y

$\displaystyle \Rightarrow x = 3^y + 1$

$\displaystyle \Rightarrow x - 1 = 3^y$

$\displaystyle \Rightarrow \log_3 (x - 1) = \log_3 3^y$

$\displaystyle \Rightarrow \log_3 (x - 1) = y$

thus, $\displaystyle g^{-1}(x) = \log_3 (x - 1)$

3. Originally Posted by Legal Eagle
Ive been having a bit of trouble with this one.

Find the Inverse.

$\displaystyle g(x)= 3^x + 1$

I think that the answer is this but Im not sure

$\displaystyle g^-1(x)=\log_{x}3 + 1$

$\displaystyle g^-1(x)=\log_{x}3 - 1$

Thanks for any and all help given.
Now you should be able to see that that is not the inverse. Using the function you defined is

$\displaystyle g\left(g^{-1}(x)\right)=x$?

4. Originally Posted by Mathstud28
Now you should be able to see that that is not the inverse. Using the function you defined is

$\displaystyle g\left(g^{-1}(x)\right)=x$?
indeed! thank you for that input. however, i should stress, it is important to check that $\displaystyle g^{-1}(g(x)) = x$ as well