# Thread: Finding the period (n)

1. ## Finding the period (n)

Hello everyone, this I'm sure is not really 'Advanced', but any help is appreciated.

I'm trying to find how many years it would take a value to triple given an effective interest rate of 20%. n being the number of years.

$100(1.20)^n=$300

How can I solve this problem algebraically?

Any help is appreciated.

2. Originally Posted by matius
Hello everyone, this I'm sure is not really 'Advanced', but any help is appreciated.

I'm trying to find how many years it would take a value to triple given an effective interest rate of 20%. n being the number of years.

$100(1.20)^n=$300

How can I solve this problem algebraically?

Any help is appreciated.
This isn't Advance Algebra at all.

You are trying to solve for n. To do so, you need to know what logorithms are. I'm hoping you do:

100(1.20)^n = 300
1.20^n = 3
log(1.20^n) = log(3)
n*log(1.20) = log(3)
n = log(3)/log(1.20)

Plug this into a calculator to find what it equals.

By the way, if you don't understand any (or all) of the steps that I just did, say so and I can explain.

3. lol, I knew that... just wanted some pity points.

Thanks very much, it gives the correct answer.

The only question I would have - and this may be one the most fundamental rules that exist in algebra... but why log(3) ends up being the numerator divided by the rate...

Is it just that you had to get n by itself by moving log(1.20) to the other side. And since it was being multiplied you have to divide on the other side?

log(1.20^n) = log(3)
n*log(1.20) = log(3)
n = log(3)/log(1.20)

4. Originally Posted by matius
lol, I knew that... just wanted some pity points.

I was a little hesitant showing the logs because you said you wanted an "algebraic method" for solving the problem, and not knowing what math class you are in I wasn't sure if you would know the properties of logs (or that logs were needed in this problem).

The only question I would have - and this may be one the most fundamental rules that exist in algebra... but why log(3) ends up being the numerator divided by the rate...

Is it just that you had to get n by itself by moving log(1.20) to the other side. And since it was being multiplied you have to divide on the other side?

log(1.20^n) = log(3)
n*log(1.20) = log(3)
n = log(3)/log(1.20)
That's exactly why. When solving for n, to get n by itself you need to divide by its coefficient, which in this case is log(1.20).

5. Originally Posted by ecMathGeek
By the way, if you don't understand any (or all) of the steps that I just did, say so and I can explain.