Question

How long does it take for $1900 to double if its invested at 9% interest compounded monthly?

Answer 1

is the interest rate per annum or per month..?

Answer 2

per month

Answer 3

A=P(1+r/n)^nt

Answer 4

So, you have A=1900(1+0.09/12)^12*2

Answer 5

ok... so 3800 = 1900(1+9/100)^n

2 = 1.09^n

take the log of both sides... base e or base 10... either is fine

ln(2) = n x ln(1.09)

n = ln(2)/ln(1.09)

you can finish it

Answer 6

n will ne the number of months

Answer 7

If it increases by 9 percent each month, then after the first month the value will be:
1900 * 1.09
Second month,
1900 * 1.09 * 1.09 = 1900 * 1.09^2
Third month,
1900 * 1.09^3
nth month,
1900 * 1.09^n

You wanna know when that will equal 3800 (double 1900) so you set it equal to it.

3800 = 1900 * 1.09^n
Divide by 1900.
2 = 1.09^n
Take the base 1.09 logarithm of both sides (refresh yourself on log identities if you don't know how to do this part).
n = 8.04323173 so approx 8 months.

To check, 1900 * 1.09^8 = 3785, so you're good.

Answer 8

This is where I left off at t=log2/12log1.0075

Answer 9

Where did the 1.0075 come from?

Answer 10

It was part of the formula (1+0.09/12)

Answer 11

And since the given rate is per month, not per year, the 12 shouldn't be there.

Answer 12

Does my answer make sense, though?

Answer 13

I was trying to figure out how to put the logarithms into my calculator

Answer 14

jewel... reading your answer.. it appears that you have used a monthly rate of 0.75% per month...

Answer 15

Logarithms can have different bases, and calculators usually only support 2 (e, aka natural log ln, and 10). You only need one base to calculate any log, though.

I won't prove this, but:
log_b a = (log_x a)/(log_x b) where x is ANY number. So, you'd take the base whatever log of a and divide that by the base whatever log of b.

Answer 16

this will alter any of the above answers, such as mine that has used 9%

Answer 17

The rule of 72 (divide 72 by the interest rate to get an approximation of doubling time) says about eight years. Lets to the more exact method to see how close the thumbnail estimate is.....

The formula is \[A(t)=2P=P(1+\frac{r}{n})^{nt}\]P is principal, r is rate, n is the number of compoundings annually, and t is the time. Plugging in for what we have in this problem...\[3800=1900(1+\frac{.09}{12})^{12t} \implies2=(1.0075)^{12t}\implies \ln(2)=12t*\ln(1.0075)\]\[\implies \frac{\ln(2)}{12\ln(1.0075)}=t \approx 7.73~years\]It turns out that the thumbnail estimate is pretty close.

Answer 18

Yes, that was the right answer.