Question

Accounts A and B both start out with $800. If Account A earns $110 per year and Account B earns 3% of its value each year, when will Account B have more money than Account A?

Answer 1

Let's start by writing out the equation for the value of each account after t years:

Account A: 800 + 110t
Account B: 800(1 + 0.03)^t

We want to find the value of t for which Account B has more money than Account A:

800(1 + 0.03)^t > 800 + 110t

We can solve this inequality using logarithms:

ln(800(1 + 0.03)^t) > ln(800 + 110t)
ln(800) + t ln(1 + 0.03) > ln(800 + 110t)
t ln(1.03) > ln(800 + 110t) - ln(800)
t > (ln(800 + 110t) - ln(800)) / ln(1.03)

Using a calculator to evaluate the right-hand side of the inequality, we find:

t > 24.5

Therefore, Account B will have more money than Account A after 25 years (since we can't have a fractional number of years).