3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 ;). $\endgroup$ – Chinny84 Dec 12 '14 at 14:31
2
$\begingroup$

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:

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

Related questions

Related question concerning equations of the form: $$e^z=A\frac{z-t}{z-s}$$

Lambert W function with rational polynomial

Solve $-B \ln y -A y \ln y + A y- A =0$ for $y$

Transcendental equations involving more than 2 terms

How to solve this equation for x? $0 = (x+k)e^{-(x+k)^2}+(x-k)e^{-(x-k)^2}$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.