Cancelling out a root function of y

**Srengam** · July 3rd 2011, 12:37 AM

Hi there,

I've got an equation and I need to find the value of $y$ . I've been able to get $y$ on it's own side of the equation but am stuck on the last steps. The left side of the equation is $y^{\frac{1}{4}}$ and I have then used the rule for powers to achieve $\sqrt[4]{y}$ but I don't know what I need to do now to both sides of the equation to get $y$ on it's own.

Many thanks.

**BAdhi** · July 3rd 2011, 12:46 AM

can you post the equation please?

**Srengam** · July 3rd 2011, 12:48 AM

I did think that the power of 4 would do it, but Mathcad isn't giving me $y$ on it's own.

I'm starting to think I may have made an error in Mathcad.

Would somebody please confirm if $(\sqrt[4]{y})^{4}=y$

Many thanks.

**Srengam** · July 3rd 2011, 12:52 AM

Originally Posted by BAdhi

can you post the equation please?

Hi, the original equation is

$\frac{-2}{y^{\frac{1}{4}}}=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}$

Thanks.

**Srengam** · July 3rd 2011, 01:02 AM

So far I've got

$\frac{-2}{y^{\frac{1}{4}}}=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}$

$-2=y^{\frac{1}{4}}[-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}]$

$y^{\frac{1}{4}}=\frac{-2}{-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}}$

**Srengam** · July 3rd 2011, 01:13 AM

Would the next step be

$y=[\frac{-2}{-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+\frac{1}{2}}]^{4}$

Thanks.

**elieh** · July 3rd 2011, 01:21 AM

$(\sqrt[4]{y})^{4}=y$

This is correct
A quick and easy way to check is to give y a value and check.

so
let y = 1000

$(\sqrt[4]{1000})^{4}$ put this in your calculator to check, and you should get 1000.

**Srengam** · July 3rd 2011, 02:54 AM

Thanks for confirming, I've been able to complete the problem.

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