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

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[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.
 
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You need to express the function as x(y); that is, make x the subject of the equation.
 
I am still not sure...Aren't I supposed to take the ln of both sides or something?
 
That would be a good first step, yes.
 
ok, ln y = ln e^(x^(1/4))
= ln y = x^(1/4)
= ln x = y^(1/4)
= (ln x)^4
...easy now lol.
 
frasifrasi said:
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...
 
Assuming he has already checked it, why "again"?

If f(x)= e^{x^{1/4}}[/itex] then f^{-1}(x)= (ln(x))^{4}<br /> <br /> For all positive x.<br /> <br /> f(f^{-1}(x))= e^{(ln(x))^4)^{1/4}}= e^{ln(x)}= x<br /> f^{-1}(f(x))= (ln(e^{x^{1/4}}))^4= (x^{1/4})^4= x<br /> <br /> Looks good to me. Of course, both functions have domain and range &quot;all positive numbers&quot;.
 
HallsofIvy said:
Assuming he has already checked it, why "again"?

If f(x)= e^{x^{1/4}}[/itex] then f^{-1}(x)= (ln(x))^{4}<br /> <br /> For all positive x.<br /> <br /> f(f^{-1}(x))= e^{(ln(x))^4)^{1/4}}= e^{ln(x)}= x<br /> f^{-1}(f(x))= (ln(e^{x^{1/4}}))^4= (x^{1/4})^4= x<br /> <br /> Looks good to me. Of course, both functions have domain and range &quot;all positive numbers&quot;.
<br /> <br /> All right, I see the intention of the original post. The inverse was calculated as:<br /> x=f^{-1}(y)<br /> and the symbols were switched. My mistake, sorry.
 
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.
 
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