Suppose $2,400 is invested in an account at an annual interest rate of 6.1% compounded
continuously. How long (to the nearest tenth of a year) will it take the investment to double in
size?

Question

Suppose $2,400 is invested in an account at an annual interest rate of 6.1% compounded
continuously. How long (to the nearest tenth of a year) will it take the investment to double in
size?

anonymous · Answer

Here we have to use the compound interest formulas . do you know it?

anonymous · Answer

yes 2400(1+0.61/12)^n*t?

anonymous · Answer

so this is the formula for interest and principal + interest = amount

We have to find the time in which the principal gets double i.e. amount is 2 x 2400 = 4800

anonymous · Answer

$$e^{.061t}=2$$ solve for $t$ in two steps

anonymous · Answer

take the log, get 
$$.061t=\ln(2)$$ divide by $.061$ to get 
$$t=\frac{\ln(2)}{.061}$$

anonymous · Answer

forget the 24000 it is a red herring, has nothing to do with the problem
doubling time is doubling time, no matter how much money you start with

anonymous · Answer

also it is continuous compounding, so you use 
$$A=P_0e^{rt}$$

anonymous · Answer

its just going to continue

anonymous · Answer

A = P + I
4800 = 2400 + 2400(1+6.1/100)^t? 
2400 = 2400(1+6.1/100)^t
t = 12 after rounding

anonymous · Answer

yes it does not matter whatever be the principal @satellite73 is all correct but i showed u to make formulas clear