How to get $y(x)$ out of $y = \sqrt[3]{xa(\log_{10}(yb)-c)^2}$ I am trying to calculate the speed of a boat given the power being delivered by the motor. Unfortunately, the friction coefficient is speed dependent and is inside a $\log_{10}$. Is it possible to solve this algebraically?
$y$ is the speed, $x$ is the power, $a$, $b$ and $c$ are constants.
$y = \sqrt[3]{xa(\log_{10}(yb)-c)^2}$
I will try to solve it until I cannot do anymore:
$y^3 = xa(\log_{10}(yb)-c)^2$
$y^3 = xa(\log^2_{10}(yb)-2\log_{10}(yb)c+c^2)$
Now what?
 A: Why not doing a Taylor approximation to linearize the problem?
A: For an exact solution you need to resort to the Lambert function $W$.
Your equation can be rewritten as
$$y=(p\ln(y)+q)^{2/3},$$ where $p,q$ are simple functions of $x$.
Then with $y=z^{2/3}$,
$$z=\frac23p\ln(z)+q.$$
Now let $z=-\frac23pt$ so that
$$-\frac23pt=\frac23p\ln(-\frac23pt)+q,$$
$$-t=\ln(-\frac23p)+\ln(t)+\frac{3q}{2p},$$
$$t+\ln(t)=-\frac23p\ln(-\frac23p)-\frac{3q}{2p},$$
$$te^t=\exp\left(-\frac23p\ln(-\frac23p)-\frac{3q}{2p}\right),$$
$$t=W\left(\exp\left(-\frac23p\ln(-\frac23p)-\frac{3q}{2p}\right)\right),$$
$$z=-\frac3{2p}W\left(\exp\left(-\frac23p\ln(-\frac23p)-\frac{3q}{2p}\right)\right),$$
$$y=\left(-\frac3{2p}W\left(\exp\left(-\frac23p\ln(-\frac23p)-\frac{3q}{2p}\right)\right)\right)^{3/2},$$
$$y=\left(-\frac3{2\frac{\sqrt{ax}}{\ln(10)}}W\left(\exp\left(-\frac23\frac{\sqrt{ax}}{\ln(10)}\ln\left(-\frac23\frac{\sqrt{ax}}{\ln(10)}\right)-\frac{3\left(\sqrt{ax}\ln_{10}(b)-c\right)}{2\frac{\sqrt{ax}}{\ln(10)}}\right)\right)\right)^{3/2}.$$
Sorry, there isn't simpler.
