Solving Inverse Function f(x) = (3 - e^(2x))^(1/2)

[Miike012]

f(x) = (3 - e^(2x))^(1/2)

y^2 = 3 - e^(2x)

-(y^2 - 3) = e^(2x)

ln(-(y^2 - 3) ) / 2 = x

What am i doing wrong?
In the back of the book it says...
ln(-(x^2 - 3) ) / 2 = y

 

[eumyang]

f(x) = (3 - e^(2x))^(1/2)

[or y= (3 - e^(2x))^(1/2)]
Here you should switch the x and y, and then solve for y. The result will give you the inverse. So instead of
y^2 = 3 - e^(2x)
you should have
x^2 = 3 - e^(2y)
... and so on.

 

[HallsofIvy]

As eumyang said, it is just a matter of notation. We typically write a function as "y= f(x)", not "x= f(y)". But, of course, as long as f is the same function "y= f(x)", "x= f(y)", or "b= f(a)" all refer to the same function.

 

[Miike012]

Knowing that log functions are the inverse of exponential functions, I thought all I had to do was find the log of the expo? This is obviously incorrect for me to do?

 

[wukunlin]

if:
$$x\rightarrow f \rightarrow y$$
then:
$$y\rightarrow f^{-1} \rightarrow x$$

which is why your notation is the opposite, in the answer of your book they redefined the inverse function to be a function of x