Question

the amount of money in an acount with countinously compounded interest is given by the formula A=pe^rt , where p is the principle, r is the anual interest rate, and t is the time in years. calculate to the nearest tenth of a year how long it takes for an acount of money to double if interest is compounded continusly at 7.5%

Answer 1

You want the money in the account to double, which means \(A=2p\):
\[2p=pe^{rt}\\
2=e^{rt}\]
The rate is 7.5%, or \(r=0.075\):
\[\large2=e^{0.075t}\\
\large\ln 2=\ln\left(e^{0.075t}\right)\\
\large\ln 2=0.075t\ln e\\
\large\ln 2=0.075t\\
t=\cdots\]

Answer 2

8.2 years?

Answer 3

9.2 Years?

Answer 4

Or you could use this formula:
Years = {log(total) -log(Principal)} ÷ log(1 + rate)
where you would make total =2 and principal =1
Years = [log(2) - log(1)] /log (1.075)
Years = [0.3010299957 -0] / 0.0314084643
Years = 0.3010299957 / 0.0314084643
Years = 9.5843589566

and I found the formula here:
www.1728.org/compint2.htm

Answer 5

it is definalty 9.2 years correct?

Answer 6

Natural log of 2 = 0.69314718056
0.69314718056 = .075*t
t = 0.69314718056 / .075
t = 9.2419624075

(guess I made a mistake in my own calculations)

Answer 7

A = Pe^(r)(t)
2 = e^0.075t
ln2 = 0.075t
0.6931 = 0.075t
t = 9.242 years

Answer 8

That seems right

Answer 9

thank you wolfman!!! lol

Answer 10

thanks :-)

Answer 11

I found out what I did wrong. I did not use a continuously compounded interest rate and just used an annual rate. (I must read the questions more carefully).