Solve $0 = ax + b \cdot \log(1 + cx)$ for $x$?

Question

I am facing a physics problem that requires me to solve $$ 0 = -gt - \frac{F}{\alpha}\cdot \log\left( 1 - \frac{\alpha t}{M+m} \right) $$ with $M=5~kg$, $m=10~kg$, $\alpha = 1 kg/s$ and $F = 0.9(M +m)g$. Via computer trial and error I was able to find that $t = 0$ and $t \approx 2.9$ are solutions. However, how can I find the solutions on a mathematical way? Since the parameters are not necessary for the mathematics, I broke it down to this for you:

$$ 0 = ax + b \cdot \log(1 + cx) $$

Note: In the exercise following hint is given: Note that $\alpha t \ll M+m$. It would be helpful already, if you could help me use that hint. Because that hint means (from my point of view) that $\frac{\alpha t}{M+m} \approx 0$. However, it would follow $\log(1 - 0) = 0$ and from that $t = 0$. This is one solutions, but how to find the second (more important) solution?

Answer 1

From your parameters, we see that $a\neq 0$, $b\neq 0$ and $c\neq 0$.

The solution of your problem is given by using the Lambert $W$ function

$$x = \dfrac{b}{a}W\left[\dfrac{a}{bc}\exp\left(\dfrac{a}{bc}\right) \right]-\dfrac{1}{c}$$

Answer 2

$$0=ax+b\ln(1+cx) \tag 1$$ Note : If $\quad 0=ax+\beta\log_{10}(1+cx)\quad\to\quad b=\beta\ln(10)$

$\ln(1+cx)=-\frac{a}{b}x\quad\to\quad 1+cx=e^{-\frac{a}{b}x}\quad\to\quad (1+cx)e^{-\frac{a}{bc}}=e^{-\frac{a}{b}x-\frac{a}{bc}}$

$\frac{bc}{a}(\frac{a}{b}x+\frac{a}{bc})e^{-\frac{a}{bc}}=e^{-\frac{a}{b}x-\frac{a}{bc}}$

$ (\frac{a}{b}x+\frac{a}{bc})e^{\frac{a}{b}x+\frac{a}{bc}}=\frac{a}{bc}e^{\frac{a}{bc}}$

The solution of $\quad Xe^X=Y\quad$ is $\quad X=W(Y)\quad$ Lambert $W$ function. http://mathworld.wolfram.com/LambertW-Function.html

With $X=(\frac{a}{b}x+\frac{a}{bc})$ and $Y=\frac{a}{bc}e^{\frac{a}{bc}} \quad\to\quad\frac{a}{b}x+\frac{a}{bc}=W\left(\frac{a}{bc}e^{\frac{a}{bc}} \right)$

The solution of the equation $(1)$ is : $$x=-\frac{1}{c}+\frac{b}{a}W\left(\frac{a}{bc}e^{\frac{a}{bc}} \right)$$

The solution cannot be expressed with a finite number of elementary functions. Many equations requires some special functions to have the solution expressed on closed form. This is the case of Equation $(1)$ which solving involves the special function $W$.

A paper for general public about special functions : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

If you don't want to use a special function, approximate solutions are commonly computed in using numerical calculus. That is the method recommended in your case insofar the numerical values of the parameters are known.

Answer 3

Using Taylor series for $\ln(1+x)$ plugging $x=-t$ we get $-x-\frac{x^2}2-\frac{x^3}{3}+\cdots$ $$0=-gt-\frac{F}{\alpha}\cdot(-\frac{\alpha t}{M+m}-\frac{\alpha^2t^2}{2(M+m)^2}-\cdots)$$ Now the more terms of the taylor expansion you take the better the approximation, since we can solve a quadratic and we know that $t=0$ is a solution we can take three terms. $$gt=F(\frac{t}{M+m}+\frac{\alpha t^2}{2(M+m)^2}+\frac{\alpha^2t^3}{3(M+m)^3})$$ $$gt=0.9g(M+m)(\frac{t}{M+m}+\frac{\alpha t^2}{2(M+m)^2}+\frac{\alpha^2t^3}{3(M+m)^3})$$ Since $t=0$ zero is a solution for $t\neq 0$ we divide by $gt$ to get $$1=0.9+\frac{0.9t}{2(M+m)}+\frac{0.9t^2}{3(M+m)^2}$$ From this $t\approx 2.94727$ (because the other solution is negative). Since $\frac{\alpha t}{M+m}\approx 0.2$ is not small enough the approximation isn't good enough, in case you can solve a $3$rd degree polynomial then taking $4$ terms gives approximately $t\approx 2.904$, larger degree of polynomial equals better approximation.