- #1
illjazz
- 59
- 0
Homework Statement
If [tex]g(x) = 3 + x + e^x[/tex], find [tex]g^{-1}(4)[/tex]
Homework Equations
Not sure. The log laws don't seem to apply. Probably laws/rules related to the number e.
The Attempt at a Solution
So I know the whole process and technically have this solved, but not because I understand how to do it but because I have it written down from a short discussion in class. It goes like this:
[tex]g(x)=3+x+e^x[/tex]
[tex]y=3+x+e^x[/tex]
[tex]y-3=x+e^x[/tex]
And this is where I get stuck. How do I "factor" (x+e^x)? Alas, what I have continues:
[tex]2x=ln(y-3)[/tex]
[tex]x=\frac{ln(y-3)}{2}[/tex]
Then,
[tex]g^{-1}(x)=y=\frac{ln(x-3)}{2}[/tex]
and
[tex]g^{-1}(4)=y=\frac{ln(4-3)}{2}[/tex]
[tex]=\frac{ln(1)}{2}[/tex]
[tex]=\frac{0}{2}[/tex]
[tex]=0[/tex]
What I don't understand is how to get from (e^x + x) to 2x.
TIA