10 points for someone who can do this algebra II question?

Question

Suppose $7,500 is invested at an annual interest rate of 5.9% compounded countinuously. Assume that there are no other deposits or withdrawals in this account.
a. What is the amount in the account after 16 years?
b. How long would it take for the amount in the account to quadruple?
Please show work and explain how you got the answers. Thank you!!!

Answer 1

a. amount = 7.5k * e^(0.059 * 16) = 19,276.81
b. 30 k = 7.5k * e^(0.059 * T)
4 = e^(0.059*T)
ln 4 = 0.059 * T
T= ln 4 / 0.059 = 23.5 years

Answer 2

This is an exponential problem.

a.

Amt = prin e^(it)

Amt = 7500 e^(0.059*16)

Amt = 7500 * 2.57 = $19,276.81

b.

4 = e^(0.059*t)

take Ln of both sides

Ln 4 = 0.059t

t = 23.5 yrs to quadruple
.

Answer 3

the equation for compounding continuously is:
I = Pe^(rt)

I = final amounts
P = principles or initial amount
r = rate (remember to convert to decimial)
t = time in year

I = 7500e^(.059 * 16)
I = $19,276.81

b) at that time, the final amount of money will be 4 times the initial amout
7500 x 4 = $30,000

30,000 = 7500e^(.059t)
4 = e^(.059t)
ln(4) = .059t
t = 23.5 years

Answer 4

Oops, forgot it's continuous, ignore the rest of this. This does not include Euler's number, and is wrong.


n=time, 1n=1 year
a). $18769.076
b). 24.183 years

Work:

f(n)=7,500 * (1.059^n)

a). n=16, f(n)=7,500 * (1.059^16)= $18769.076

b). f(n)=4*7500=30000=7500*(1.059^n)
30,000/7500=4=1.059^n
4=1.059^n
log(4)/log(1.059)=n=24.183 years