Let's say I bought a property for 80,000 that I will be renting for 1,200 a year with a 5% yearly increase from the previous year.
Let's call y the number of years required to get my money back. Of course we're ignoring all other expenses and missed payments.
So far everything is simple. 80000=sum(1200*1.05^y) and we solve for y.
Solve for y with Wolfram Alpha
Let's assume now that the increase rate is unknown and call it t. What I'm trying to do is to plot the graph that gives y in terms of t. I want to be able to read information like. y=20 when t=2, y=12 when t=5... (Those examples must be wrong).
I tried this on Wolfram Alpha but still can't get the graph that I'm looking for. I'm very novice with the tool.
You have $$80000=1200\sum_{x=0}^y (1+t)^x$$
It exists a close formula for the sum, because it is the partial sum of a geometric series.
$$80000=1200\cdot \frac{(1+t)^{y+1}-1}{t}$$
$\frac{200}3t=(1+t)^{y+1}-1$
$\frac{200}3t+1=(1+t)^{y+1}$
Taking logs on both sides
$\ln\left(\frac{200}3t+1 \right)=(y+1)\cdot \ln(1+t)$
$$y(t)=\frac{\ln\left(\frac{200}3t+1 \right)}{\ln(1+t)}-1$$
For $t=0.05$ you get y=29.0539
The plot on wolfram alpha looks like