How to get $y(x)$ out of $y = \sqrt[3]{xa(\log_{10}(yb)-c)^2}$

Question

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?

Answer 1

Why not doing a Taylor approximation to linearize the problem?

Answer 2

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.