# Homework Help: Inverse Function

1. Oct 14, 2007

### momogiri

Question:
If g(x) = 5 + x + e^x, find g^-1(6) [inverse of g, not g to the power of -1]

Attempted:
So I first substitued g(x) to y

So y = 5 + x + e^x
then I tried isolating the x
So y - 5 = x + e^x
Then I applied ln to both sides
ln(y) - ln(5) = ln(x) + ln(e^x)
Due to the log rules, I can take down the x from ln(e^x)
So it becomes ln(y) - ln(5) = ln(x) + x(ln(e))
And since ln(e) = 1, that means
ln(y) - ln(5) = ln(x) + x

Ok. So now I'm stuck.
Where should I go from here?? I know I'm missing something extremely crucial here, but I'm really not seeing it :(

2. Oct 14, 2007

### dt_

You applied the ln incorrectly. You'd be getting ln (y-5), not ln y - ln 5.

See if you can go from there

Anyway, the idea for finding an inverse function like this is, replace all the y with x, and replace all the x with y. Then solve for the new y, which is the inverse function. Then plug in the given x.

3. Oct 14, 2007

### momogiri

Oh, really? XD
So if it's ln(y - 5) then x and e^x would be ln(x + e^x) right?
I.. don't know where to go from there.. :(

4. Oct 14, 2007

### Gib Z

You could think of the question as saying: What value of x would make $$5+x+e^x=6$$?

In other words, find x, if $$x+e^x = 1$$.

Now, you can either take the log of both sides here, but really its just easy to see that its 0.

5. Oct 14, 2007

### momogiri

Oh wow.. so I was making this question harder than it seems XD
Thanks so much, Gib Z!! :D
Brownie points for you :D :D