I am struggling with this problem: At $x=0$ I own $b$ units. Every year I deposit $a$ units. The bank pays an interest rate of $c$ every year. After how many years I will own $d$ units?
$$(ax+b)*e^{cx} = d$$
How am I able to solve this for $x$?
I know that $$ax+b = d \Rightarrow x = {d-b \over a}$$ and $$e^{cx} = d \Rightarrow x ={\ln d \over c}$$ but I have no idea how to solve the first equation. Thank you for any help!
You can't, not with elementary functions. As soon as your unknown is both outside and inside exp/log functions, it's hopeless - there's no closed form in terms of elementary functions. You have to attack the problem numerically (iteration, bisection, Newton's method,...). In this case, you can express the solution through the Lambert's W function, but that's just a fancy name for a numerical procedure.
The amount you have at the beginning is $b$.
After $1$ year, you have $a+b(1+c)$.
After $2$ years, you have $a+(a+b(1+c))(1+c)$.
After $3$ years, you have $a+(a+(a+b(1+c))(1+c))(1+c)$.
etc.
After $N$ years, you have
$$b(1+c)^N+a(1+c)^{N-1}+a(1+c)^{N-2}+\cdots+a(1+c)+a$$ $$=\left(b+\frac{a}{c}\right)(1+c)^N-\frac{a}{c}$$
For a given $d$, in general you can't have integer $N$ which gives you the exact amount $d$. But you have find the integer $N$ that you amount first exceed $d$, by
$$N=\bigg\lceil \frac{\log\frac{d+a/c}{b+a/c}}{\log(1+c)} \bigg\rceil $$
If we are allowed to approximate it as a continuous time process, then we have $$\frac{dM}{dx}=a+cM$$ where $M$ is the amount of money you have at $x$.
The solution is $$M=\left(b+\frac{a}{c}\right)e^{cx}-\frac{a}{c}$$
Then you can solve $$\left(b+\frac{a}{c}\right)e^{cx}-\frac{a}{c}=d$$ to obtain $$x=\frac{1}{c}\ln\frac{d+a/c}{b+a/c}$$
