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

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Homework Help Overview

The discussion revolves around finding the inverse of the function y = e^(x^(1/4)). Participants are exploring the steps necessary to express x in terms of y.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to isolate x by taking the natural logarithm of both sides. There is uncertainty about the correct steps to manipulate the equation, with some suggesting that taking the ln is a good first step.

Discussion Status

There is a mix of attempts to clarify the process of finding the inverse. Some participants provide guidance on expressing the function in terms of x, while others question the correctness of the steps taken. Multiple interpretations of the function and its inverse are being explored.

Contextual Notes

Participants note that the function and its inverse are defined for all positive numbers, and there is some confusion regarding the notation and steps involved in the process.

frasifrasi
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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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