Inverse Exponential Function?

**Legal Eagle** · June 30th 2008, 04:15 PM

Ive been having a bit of trouble with this one.

Find the Inverse.

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

I think that the answer is this but Im not sure

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

Or would the answer be

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

Thanks for any and all help given.

**Jhevon** · June 30th 2008, 04:28 PM

Originally Posted by Legal Eagle

Ive been having a bit of trouble with this one.

Find the Inverse.

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

I think that the answer is this but Im not sure

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

Or would the answer be

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

Thanks for any and all help given.

neither is correct

$y = 3^x + 1$

for inverse, switch x and y and solve for y

$\Rightarrow x = 3^y + 1$

$\Rightarrow x - 1 = 3^y$

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

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

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

**Mathstud28** · June 30th 2008, 07:59 PM

Originally Posted by Legal Eagle

Ive been having a bit of trouble with this one.

Find the Inverse.

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

I think that the answer is this but Im not sure

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

Or would the answer be

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

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

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

**Jhevon** · June 30th 2008, 09:36 PM

Originally Posted by Mathstud28

Now you should be able to see that that is not the inverse. Using the function you defined is

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

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

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