While trying to answer this SO question I got stuck on a messy bit of algebra: given
$$
\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)
$$
I need to solve for $m$. I no longer remember enough logarithmic identities to attempt to do this by hand. Maxima can’t do it at all, and Wolfram Alpha coughs up a hairball that appears to be the zeroes of a quartic, with no obvious relationship to the original equation.
Is there a short, tidy solution? Failing that, an explanation of how WA managed to turn this into a quartic, and the quartic itself, would be ok.
Best Answer
$$\log {m}=\log {n}+\frac{3}{2} \log {\left(1+\frac{v}{m^2}\right)}$$ $$\log {m}=\log {n}+ \log {\left(\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}\right)}$$ $$\log {m}=\log {\left({n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}}\right)}$$ Exponentiate both sides: $$m=n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}$$ $$m^2=n^2\cdot\left(1+\frac{v}{m^2}\right)^3$$ $$m^2=n^2\cdot\left(1+\frac{3v}{m^2}+\frac{3v^2}{m^4}+\frac{v^3}{m^6}\right)$$ $$m^2=\frac{n^2}{m^6}\left(m^6+3v\cdot m^4+3v^2 \cdot m^2+v^3\right)$$ $$m^8=n^2 m^6+3vn^2 m^4+ 3n^2v^2 m^2+n^2v^3$$ Substitute $m^2=u$ and you will obtain a quartic expression. $$u^4-n^2u^3-3vn^2 u^2-3n^2 v^2 u-n^2v^3=0$$ This is going to be a long solution, however you can use the general formula for the solution to a quartic equation.
Wikipedia suggests using substitutions in order to solve it.
The full formula without substitution for $ax^4+bx^3+cx^2+dx+e$ is below (I cannot write it in $\LaTeX$ because it is so long).
This suggests that your solution in Wolfram Alpha is probably correct.