# inverse function

• Jan 24th 2010, 10:42 PM
windir
inverse function
If $\displaystyle f(x)=2x+ln x$, find $\displaystyle f^{-1}(2)$.
I keep getting stuck trying to get the inverse. Here's what i got:
$\displaystyle y=2x+lnx$
$\displaystyle -e^{lnx}+e^y=e^{2x}$
$\displaystyle -x+e^y=e^{2x}$
Then i don't know what to do, and i feel like i'm approaching it incorrectly.
So, i tried: $\displaystyle \frac{y}{2x}=lnx$
$\displaystyle x=e^{\frac{y}{2x}}$ and i'm still not getting anywhere.
Any help would be mucho appriciated!
• Jan 24th 2010, 10:57 PM
Jhevon
Quote:

Originally Posted by windir
If $\displaystyle f(x)=2x+ln x$, find $\displaystyle f^{-1}(2)$.
I keep getting stuck trying to get the inverse. Here's what i got:
$\displaystyle y=2x+lnx$
$\displaystyle \color{red}-e^{lnx}+e^y=e^{2x}$
$\displaystyle -x+e^y=e^{2x}$
Then i don't know what to do, and i feel like i'm approaching it incorrectly.
So, i tried: $\displaystyle \frac{y}{2x}=lnx$
$\displaystyle x=e^{\frac{y}{2x}}$ and i'm still not getting anywhere.
Any help would be mucho appriciated!

by inspection, $\displaystyle f^{-1}(2) = 1$ since $\displaystyle f(1) = 2(1) + \ln 1 = 2 + 0 = 2$

(by the way, what you did is incorrect. from the line in red)
• Jan 24th 2010, 11:26 PM
windir
ah thank you! obvious stuff usually trips me up!
• Jan 24th 2010, 11:30 PM
Jhevon
Quote:

Originally Posted by windir
ah thank you! obvious stuff usually trips me up!

you'd be surprised how common this vice is. take care.