Solve x from y=e^(x^(1/4)): Find Inverse

[frasifrasi]

[SOLVED] finding the Inverse...

The question asks to find the inverse of

y = e^(x^(1/4))

--> I kind of forgot how to proceed for something like this, if anyone can help me, it would be great.

 

[cristo]

You need to express the function as x(y); that is, make x the subject of the equation.

 

[frasifrasi]

I am still not sure...Aren't I supposed to take the ln of both sides or something?

 

[cristo]

That would be a good first step, yes.

 

[frasifrasi]

ok, ln y = ln e^(x^(1/4))
= ln y = x^(1/4)
= ln x = y^(1/4)
= (ln x)^4
...easy now lol.

 

[coomast]

ok, ln y = ln e^(x^(1/4))
= ln y = x^(1/4)
= ln x = y^(1/4)
= (ln x)^4
...easy now lol.

I would suggest to check this again...

 

[HallsofIvy]

Assuming he has already checked it, why "again"?

If f(x)= [tex]e^{x^{1/4}}[/itex] then $f^{-1}(x)= (ln(x))^{4}$

For all positive x.

$$f(f^{-1}(x))= e^{(ln(x))^4)^{1/4}}= e^{ln(x)}= x$$
$$f^{-1}(f(x))= (ln(e^{x^{1/4}}))^4= (x^{1/4})^4= x$$

Looks good to me. Of course, both functions have domain and range "all positive numbers".

 

[coomast]

Assuming he has already checked it, why "again"?

If f(x)= [tex]e^{x^{1/4}}[/itex] then $f^{-1}(x)= (ln(x))^{4}$

For all positive x.

$$f(f^{-1}(x))= e^{(ln(x))^4)^{1/4}}= e^{ln(x)}= x$$
$$f^{-1}(f(x))= (ln(e^{x^{1/4}}))^4= (x^{1/4})^4= x$$

Looks good to me. Of course, both functions have domain and range "all positive numbers".

All right, I see the intention of the original post. The inverse was calculated as:
$$x=f^{-1}(y)$$
and the symbols were switched. My mistake, sorry.

 

[pmbasa]

y = e^(x^(1/4))

x = e^(y^(1/4))
ln^x = y^1/4
y = (ln^x)^4

i guess you got it.. lol, i took a few seconds of my surfing time to solve it when you already got it.