You invest $15,000 into an account that pays 10% interest compounded continuously. How long must the money be left in the account for it to grow to $110,835.84?

Answer 1

For a continuous compounding A=Pe^{rt} where A is the balance, P is principal, t is years and r is annual interest rate (in decimal form).

In this case 110,835.84=15000 e^(0.1t)

You have to find t which means you have to solve for t

110835.84/15000=e^(0.1t)

7.3890=e^(0.1t)

apply ln on both side we have:

ln(7.3890)=0.1t

1.999=0.1t

t=19.99

Therefore after approximately 20 years your investment of 15000 will grow to 110,835.84

Answer 2

The equation for continuously compounded interest is y=Pe^rt.

In this case, the principal is $15,000, rate is 0.1 and the total amount is $110,835.84.

So, you have:

110835.84 = 15000e^0.1*t.

Divide both sides by 15000

7.39 = e^0.1*t

The inverse of this exponential function is natural log.

ln 7.39 = 0.1*t

2 = 0.1t

t = 20 years