Finding inverse of a function involving ln.

[sp09ta]

1. Plot f(x)=5+ln(-2x+3), and its inverse.



2. I know the inverse of ln(x) is exp(1)^(x)., and to find the inverse of a function, solve for x then swap the x and y.



3. y=5+ln(-2x+3)
y-5=ln(-2x+3)
exp(1)^(y-5)=(-2x+3)
(-1/2)*[exp(1)^(y-5)-3]=x

y=(-1/2)*[exp(1)^(x-5)-3]

However when I plot these functions using maple, I notice that they are definitely not inverse... :( what am i doing wrong?

 

[CompuChip]

Looks good (although it beats me why you write exp(1)^p instead of just exp(p) or e^p).
How can you "see" that they are not inverses?
If you plug x = (-1/2)*[exp(y-5)-3]=x into 5+ln(-2x+3), you get y right?

 

[sp09ta]

Oh, I guess my theory on the geometry of inverses was a little off. I write exp(1)^p because its the maple command that I was using. Thanks for your comment!

 

[CompuChip]

Hehe, you're welcome.
Sometimes all you need is a little reminder on the definitions :)