What is the equation for the account given a deposit and a gain every year then finding the balance of this account after an 18 year period?

Question

A principal of $3500 is deposited in a savings account that gains 4.9% annually.

I’m supposed to write an equation to model the situation.

And figure out how much money will be in the

account after 18 years.

Then find out which year the account will exceed $6000?

Answer 1

Hi I'd love to help you understand this question!

You are trying to find the equation to determine the value present in a savings account after 18 years given an initial deposit of $3500 and an interest rate of 4.9%.

First you know this:

Initial deposit: P = 3500

Interest Rate: r = 4.9%

Time: t= 18 years

Interest is compounded annually

The formula needed is A(t) = P(1 + r)^t

when this is plugged in: A(t) = 3500(1 + 0.049)^18 You would include 0.049 instead of 4.9 because it needs to be in decimal format rather than percentage format.

A(18) = 3500(1.049)^18 = 3500(2.36569519342) ≈ 8279.933

This means that the savings account will have $8279.933 after 18 years given the interest rate and initial deposit.

Part 2) What year does the savings account exceed $6000

The equation would be 3500(1.049)^t > 6000

(This is because we are using the same standard interest formula, but trying to isolate t when the value of the savings account is greater than 6000)

3500(1.049)^t > 6000

(1.049)^t > 6000/3500 =1.7143

(1.049)^t > 1.7143

Take the Log of both sides to isolate t

t x log(1.049) > log(1.7143)

t > log(1.7143)/log(1.049) ≈ 11.5

t > 11.5

We round up to 12 and say that we exceed $6000 in the saving account during the 12th year

I hope this helps!