# help for my son

• June 14th 2013, 01:23 AM
help for my son
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
• June 14th 2013, 01:38 AM
emakarov
Re: help for my son
As far as I know, Y(X) cannot be expressed in elementary functions in general.
• June 14th 2013, 02:13 AM
MINOANMAN
Re: help for my son
Mark.. Emakarov is right

you cannot express y as an algebraic expression ...since it is trancedental.
such type of equations can be solved only by approximation or graphically .
• June 14th 2013, 02:21 AM
$X^y= e^{ln(X^Y)= e^{Y ln(X)}$ so the equation can be written as $Ce^{Y ln(X)}= PY$ or $\frac{C}{P}= Ye^{-Yln(X)}$. Let Z= -Yln(X). Then $Y= -Zln(X)$ and $e^{-Yln(X)}= e^Z$ so the equation becomes $\frac{C}{P}= -Zln(X)e^Z$ and then $-\frac{C}{PZln(X)}= Ze^Z$.
Now apply "Lambert's W function" to both sides. "Lambert's W function" is defined as the inverse function to $f(x)= xe^x$ so $W(Ze^Z)= Z= W\left(-\frac{C}{PZln(X)}\right)$. Of course, "Lambert's W function" is NOT an "elementary function" so emakarov is completely correct.