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How do I solve the equation $\displaystyle x=ay^2(by)^{\frac 1y}$ for $y$, where $a$ and $b$ are constants? I've been trying to manipulate this into a form on which I can use the Lambert W function, but I don't know whether this is possible or if so how to do it.

Sorry about the title, I couldn't think of anything more accurate without putting the entire equation in the title and I didn't want to just say "solving equation" or some such.

Avi
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  • Trying the sub $$by = \frac{1}{v}$$ we can obtain $$ u^2 (u^u)^{-b} = u^{2-bu}= \frac{a}{b^2}\frac{1}{x} $$ Maybe this can help..atm I can not see it ;). – Chinny84 Dec 12 '14 at 14:31

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Taking the logarithm of: $\displaystyle x=ay^2(by)^{\frac 1y}$ we obtain: $$\log(x)=\log(a)+2\log(y)+{\log(by) \over y}$$ placing $z=\log(by)$: $$\log(x)=\log(a)+2z+({\log(b)+z})e^{-z}$$ $$e^{-z}={\log(x/a)-2z \over {\log(b)+z}}$$ or: $$e^{z}=\frac{\log(b)+z }{\log(x/a)-2z }$$ We know that mixed exponential/bilinear equations can be solved by the extended Lambert function $W_r(x)$, which can be represented by the following Lagrange inverting series:

$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$

which is the solution of:

$$e^z=A\frac{z-t}{z-s}$$

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References

On the generalization of the Lambert $W$ function with applications in theoretical physics, http://arxiv.org/abs/1408.3999

[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337. http://www4.ncsu.edu/~ces/pdfversions/68.pdf

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giorgiomugnaini
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  • I don’t follow your first substitution. From the equation $\log x= \log a +2\log y + \frac{\log(by)}{y}$, substituting $z=\log (by)$ and solving for $y$ in terms of $z$ I got $y=\frac{10^z}{b}$. Plugging into the original equation, I got $\log x = \log a + 2 \log y + \frac zy$ and then $\log x = \log a + 2(z- \log b) + \frac zy$ and then $\log x = \log a + 2z - 2\log b + bz10^{-z}$ and then $10^{-z}= \frac{\log(\frac xa)-2z+2\log b}{bz}$. Also, I notice that your equations use “log” but you introduced base e. Isn’t that incorrect? A log with an unspecified base has base 10. Ln has base e. – Radial Arm Saw Jul 20 '20 at 01:54