If you deposit P dollars into a bank account paying an annual interest rate r, with n interest payments each year, the amount A you would have after t years is A=P(1+r/n)^nt. Kevin places $100 in a savings account earning 6% annual interest, compound quarterly. If Kevin adds no more money to his account, how long will it take his money to double?

Question

If you deposit P dollars into a bank account paying an annual interest rate r, with n interest payments each year, the amount A you would have after t years is A=P(1+r/n)^nt. Kevin places $100 in a savings account earning 6% annual interest, compound quarterly. If Kevin adds no more money to his account, how long will it take his money to double?

Answer 1

P = Po(1+r)^n.

r = (6%/4)/100% = 0.015 = Quarterly %
rate expressed as a decimal.

n = 4comp/yr * t = 4t.

P = 100(1.015)^4t = 200
(1.015)^4t = 200/100 = 2.00
4t*Log1.015 = Log2
4t = Log2/Log1.015 = 46.556
t = 11.64 Yrs.

Answer 2

To determine how long it will take for Kevin's money to double, we need to find the value of t in the formula A = P(1 + r/n)^(nt).

In this case, Kevin initially places $100 into the savings account. The annual interest rate is 6%, or 0.06, and the interest is compounded quarterly, meaning n = 4.

Let A represent the doubled amount, which is $200. We can now rewrite the formula as:

$200 = $100(1 + 0.06/4)^(4t)

Now we need to solve for t:

(1 + 0.06/4)^(4t) = 2

To solve this equation, we can take the natural logarithm (ln) of both sides:

ln[(1 + 0.06/4)^(4t)] = ln(2)

Using the logarithmic property ln(a^b) = b * ln(a), we can simplify the equation further:

4t * ln(1 + 0.06/4) = ln(2)

Now, we can isolate the variable by dividing both sides of the equation by 4 * ln(1 + 0.06/4):

t = ln(2) / (4 * ln(1 + 0.06/4))

Calculating the right side of the equation will give us the value of t, which represents the number of years it takes for Kevin's money to double in the savings account.