Cancelling out a root function of y

Hi there,

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

Many thanks.

Re: Cancelling out a root function of y

can you post the equation please?

Re: Cancelling out a root function of y

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

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

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

Many thanks.

Re: Cancelling out a root function of y

Quote:

Originally Posted by

**BAdhi** can you post the equation please?

Hi, the original equation is

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

Thanks.

Re: Cancelling out a root function of y

So far I've got

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

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

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

Re: Cancelling out a root function of y

Would the next step be

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

Thanks.

Re: Cancelling out a root function of y

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

This is correct

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

so

let y = 1000

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

Re: Cancelling out a root function of y

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