SOLUTION: How do i find the inverse function of: f(x)=ln√(e^x - 2) I started it like this: f(x)=y <=> y=ln√(e^x-2) <=> e^y = √(e^x-2) <=> e^2y = e^x - 2 <=> e^x = e^2y+2 <=> ???

Algebra.Com
Question 1163140: How do i find the inverse function of: f(x)=ln√(e^x - 2)
I started it like this:
f(x)=y <=> y=ln√(e^x-2) <=> e^y = √(e^x-2) <=> e^2y = e^x - 2 <=> e^x = e^2y+2 <=> ???
But I couldn't solve it till the end so i would appreciate any kind of help!

Found 3 solutions by Edwin McCravy, Theo, greenestamps:
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!


You were almost there!



 

  

 





Now interchange x and y



Replace y by f-1(x)

f-1(x) = ln(e2x+2)

Edwin


Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
this was a beauty, but i think i have it, even though it's not what i would have thought.

normally, when you solve for the inverse function, you do the following:

replace y with x and x with y and then solve for y.

for example, consider y = ln(x)
replace x with y and y with x to get x = ln(y)
this is true if and only if y = e^x.
the inverse function of y = ln(x) is y = e^x. *****
that can be seen in the following graph.



we'll do the same procedure with your problem.
your equation is y = ln(sqrt(e^(x-2))
replace x with y and y with x to get x = ln(sqrt(e^(y-2))
this is true if and only if e^(y-2) = e^x
this is true if and only if y-2 = x
solve for y to get y = x + 2
your inverse equation is y = x + 2 *****
that can be seen in the following graph.



you will notice that both the graph of the original equation and the graph of the inverse equation are straight lines.

i figured if the original equation was a straight line and was shown in the form of a natural log function, then the inverse function, being a straight line, could also be shown in the form of a natural log funcion.

i ran out of time to explain this fully, but the results of my investigation was that the inverse function could also be shown in the form of a natural log function.

i got the following:

the original function is y = ln(sqrt(e^(x-2))
the inverse function is y = ln(e^(2x+2))
that can be shown in the following graph.




take it for what it's worth.
your inverse function is either:
y = 2x + 2, or:
y = ln(e^(2x+2))

both of these can be seen on the graphs.
the graphs show them to be reflections about the line y = x and (x,y) = (y,x).
both of these are indications of inverse functions.

i'm actually surprised i was able to see the relationship, but it's there, even it i can't explain if very well.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The answer from the other tutor demonstrates the standard way of finding the inverse of a function, which you were trying to do.

For many functions, it is easy to find an inverse by using the idea that an inverse function "un-does" what the function does.

Using that concept, the inverse function has to perform the opposite operations, and in the reverse order, compared to the original function.

The given function does the following to the input value:
(1) raise e to that power
(2) subtract 2
(3) take the square root
(4) take the natural log

The inverse function must then
(1) raise e to the power
(2) square it
(3) add 2
(4) take the natural log

That sequence of operations gives us the inverse function:

--> --> --> -->


RELATED QUESTIONS

Find the inverse of the logarithmic function f(x)=-ln(x-2). Steps I've done so far:... (answered by robertb)
I'm trying to find the inverse of the following function: f(x)=ln(x-2)+1. Here's what I (answered by lwsshak3)
What is the inverse of f(x)=ln(x-2)? I got y=e^x +2, but I don't think that's right...I... (answered by scott8148)
Use logarithmic differentiation to solve: f(x) =(x^(3/2)*e^(-(x^2)))/(1-e^(x)) Don't (answered by jim_thompson5910)
Morning, Please could somebody possible help with this Right its been a while since (answered by Fombitz)
Evening All, Right its been a while since i have attempted maths, and have just gone... (answered by Alan3354)
Evening All, Right its been a while since i have attempted maths, and have just gone back (answered by user_dude2008,Fombitz)
Simplify the expression... (answered by lynnlo,Alan3354)
Simplify... (answered by Alan3354)