9514 1404 393
Answer:
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.
