Can someone help me solve this for my son? I have gone blank!

I need to solve this for Y.

C.X^Y = P.Y

thanks

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- Jun 14th 2013, 01:23 AM #1

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## Re: help for my son

$\displaystyle X^y= e^{ln(X^Y)= e^{Y ln(X)}$ so the equation can be written as $\displaystyle Ce^{Y ln(X)}= PY$ or $\displaystyle \frac{C}{P}= Ye^{-Yln(X)}$. Let Z= -Yln(X). Then $\displaystyle Y= -Zln(X)$ and $\displaystyle e^{-Yln(X)}= e^Z$ so the equation becomes $\displaystyle \frac{C}{P}= -Zln(X)e^Z$ and then $\displaystyle -\frac{C}{PZln(X)}= Ze^Z$.

Now apply "Lambert's W function" to both sides. "Lambert's W function" is**defined**as the inverse function to $\displaystyle f(x)= xe^x$ so $\displaystyle W(Ze^Z)= Z= W\left(-\frac{C}{PZln(X)}\right)$. Of course, "Lambert's W function" is NOT an "elementary function" so emakarov is completely correct.