3000 dollars is invested in a bank account at an interest rate of 7 percent per year, compounded continuously. Meanwhile, 20000 dollars is invested in a bank account at an interest rate of 5 percent compounded annually. To the nearest year, when will the two accounts have the same balance?

9514 1404 393

  after 89 years

Step-by-step explanation:

For principal p, interest rate r, and number of years t, the two account balances are ...

  a = p·e^(rt) . . . . continuous compounding

  a = p(1+r)^t . . . . annual compounding

Using the given values, we have

  3000·e^(0.07t) . . . . . compounded continuously

  20000·1.05^t . . . . . . compounded annually

We want to find t so these are equal.

  3000·e^(0.07t) = 20000·1.05^t

  0.15e^(0.07t) = 1.05^t . . . . divide by 20,000

  ln(0.15) +0.07t = t·ln(1.05) . . . . take natural logarithms

  ln(0.15) = t·(ln(1.05) -0.07) . . . . subtract 0.07t

  t = ln(0.15)/(ln(1.05) -0.07) ≈ -1.8971/-0.02121 . . . . . divide by the coefficient of t

  t ≈ 89.4 ≈ 89

The two accounts will have the same balance after 89 years.

answer: the common difference is -0.10