Calculating time to 0? Quick question format:
Let $a_n$ be the sequence given by the rule: $$a_0=k,a_{n+1}=\alpha a_n−\beta$$
Find a closed form for $a_n$.
Long question format:
If I have a starting value $x=100000$ then first multiply $x$ by $i=1.05$, then subtract $e=9000$.  Let's say $y$ is how many times you do it.
Anyone know of a formula to get $e$ given $y,x,i$ and given the total must be $0$ after $y$ "turns"?
Example:
$$y=1: xi-e=0$$
$$y=2: ((xi-e)i-e)=0$$
and so on...
 A: I will change notation. Let $A$ be what you call $x$, let $P$ be what you call $e$. Let $i$ be the interest rate, which would in your concrete case be $0.05$, that is, $5\%$. I will leave your $y$ alone, though I am tempted to call it $n$.  
Let us calculate, patiently at first. 
After $1$ year we have $A(1+i)-P$.
After $2$ years we have $A(1+i)^2-P(1+i)-P$. (We multiplied the previous result by $1+i$, then subtracted $P$.)
After $3$ years we have $A(1+i)^3-P(1+i)^2-P(1+i)-P$.
Calculate a little more, perhaps writing down what we have after $4$ years, and after $5$ years.  
You will see that after $y$ years we have
$$A(1+i)^y -P\left( (1+i)^{y-1}+(1+i)^{y-2}+\cdots +1\right).$$
The finite geometric series $1+(1+i)+(1+i)^2+\cdots +(1+i)^{y-1}$ has sum $\dfrac{(1+i)^y-1}{i}$.
So setting what we have left after $y$ years equal to $0$, we get
$$P=Ai\frac{(1+i)^y}{(1+i)^y-1}.$$
A: Let's $u_0 = x$, and $u_{n+1} = iu_n - e$.
To solve this equation, let's $l= il-e$, and $v_n = u_n - l$.
Then $v_{n+1}= i v_n$ ans $v_n = i^n v_0$, so $u_n = l + i^n v_0 = l + i^n (x - l)$.
But $l = e/(i-1) = 180 000$. So $u_n = 180 000 - 80000*1.05^n$. So $u_n \leq 0$ is equivalent to
$$180 000 - 80000*1.05^n \leq 0$$
$$1.05^n \geq 180000/80000 = 2.25$$
$$n \geq 17$$
(And $u_{17} \neq 0$ !)
General case :
Let's $l = e/(i-1)$. If $i > 1$ and $x > l$, $u_n \leq 0$ is equivalent to
$$l + i^n(x-l) \leq 0$$
$$i^n \geq l/(x-l)$$
$$n \geq \frac{log(i)}{log(l) - log(x-l)}$$
So $$y = E(\frac{log(i)}{log(l) - log(x-l)}) + 1$$
