Question

4400 dollars is placed in an account with an annual interest rate of 7.5%. To the nearest tenth of a year, how long will it take for the account value to reach 33800 dollars?

Answer 1

Answer:

Step-by-step explanation:

The equation for the amount of money in an account after a certain amount is deposited and compounded after t years once per year is

$A(t)=P(1+r)^t$

Our A(t) = 33800, P = 4400, r = .075 and we are looking for t.  Filling in:

$33800=4400(1+.075)^t$ and

$33800=4400(1.075)^t$

Begin by dividing both sides by 4400 to get

$7.681818182=1.075^t$

The only way to move that t our from its current position as an exponent is to take the natural log of both sides and follow the rules for natural logs:

$ln(7.6181818182=ln(1.075)^t$

The power rule of natural logs says we can move that exponent down in front, giving us:

$ln(7.681818182)=t*ln(1.075)$

Divide both sides by ln(1.075) to get

$\frac{ln(7.61818182)}{ln(1.075)} =t$

Do this division on your calculator to get

t = 28.2 years

Answer 2

Answer: it take 28.5 years for the account value to reach 33800 dollars

Step-by-step explanation:

We would assume that the interest was compounded annually. The formula for determining compound interest is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $4400

A = $33800

r = 7.5% = 7.5/100 = 0.075

n = 1 because it was compounded once in a year.

Therefore,

33800 = 4400(1 + 0.075/1)^1 × t

33800/4400 = (1 + 0.075)^t

7.68 = (1.075)^t

Taking log of both sides, it becomes

Log 7.68 = t log 1.075

0.885 = 0.031t

t = 0.885/0.031

t = 28.5 years