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6-3: PRACTICE Part 2 Logarithms in Equations Algebra 2
Michael invests $1,000 in an account that earns a 4.75% annual percentage rate compounded continuously. Peter invests $1,200 in an account that earns a 4.25% annual
percentage rate compounded continuously. Which person's account will grow to $1,800 first?
Michael's account will grow to $1,800 after about year(s). Peter's account will grow to $1,800 after about year(s). So,
(Round to the nearest whole number as needed.)
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Answer 1

Answer: To solve this problem, we need to use the continuous compound interest formula:

A = Pe^(rt)

where A is the amount in the account, P is the initial principal, e is the mathematical constant e (approximately 2.71828), r is the annual interest rate (as a decimal), and t is the time in years.

For Michael's account, we have:

A = 1000e^(0.0475t)

For Peter's account, we have:

A = 1200e^(0.0425t)

We want to find the time it takes for each account to reach $1,800. So we can set up the following equations:

1000e^(0.0475t) = 1800

1200e^(0.0425t) = 1800

We can solve each equation for t by taking the natural logarithm of both sides and isolating t:

ln(1000) + 0.0475t = ln(1800)

ln(1200) + 0.0425t = ln(1800)

Subtracting ln(1000) or ln(1200) from both sides, we get:

0.0475t = ln(1800) - ln(1000)

0.0425t = ln(1800) - ln(1200)

Dividing both sides by the interest rate and simplifying, we get:

t = (ln(1800) - ln(1000)) / 0.0475 ≈ 10.16 years for Michael's account

t = (ln(1800) - ln(1200)) / 0.0425 ≈ 10.62 years for Peter's account

Therefore, Michael's account will grow to $1,800 first, after about 10 years (rounded to the nearest whole number).

Explanation: