Dayaan M. answered • 08/26/26
Earned A’s Twice in Precalculus | 5 Years of Tutoring Experience
In order to solve this, we start with the compound interest formula, which for interest compounded once a year looks like:
A = P(1 + r)^t
Here P is what you start with, r is the rate as a decimal, t is the number of years, and A is what you end up with. Filling in what we know, P is 500, r is 0.06, and A is 1300:
1300 = 500(1.06)^t
The goal is to solve for t, and notice that t is stuck up in the exponent. That is the whole difficulty of this problem, and it is also why this shows up in precalculus rather than algebra 1. To pull a variable down out of an exponent, we use logarithms.
First let us get the exponential part by itself by dividing both sides by 500:
2.6 = 1.06^t
Now we take the natural log of both sides. The reason this helps is the power rule for logs, which says ln(a^t) = t times ln(a). So the t comes down in front and becomes something we can actually solve for:
ln(2.6) = t times ln(1.06)
Dividing both sides by ln(1.06):
t = ln(2.6)/ln(1.06)
Running that through a calculator, ln(2.6) is about 0.9555 and ln(1.06) is about 0.0583, so t is about 16.4.
It is worth checking this back against the original. 500 times 1.06^16.4 comes out to about $1,300, so it works.
So, our final answer is that it would take about 16.4 years.
