+16 suppose $5400 is invested in an account at an annual interest rate of 7.2% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size?
+129842 We need to solve this for t
10800 = 5400e^(.072 * t) divide both sides by 5400
2 = e^(.072 * t) take the ln on both sides
ln 2 = ln e^(.072 * t) and, by a property of logs, we can write
ln 2 = (.072 * t) * ln e and since ln e = 1, we can ignore this
ln 2 = .072t divide both sides by .072
ln 2 / .072 = t = about 9.6 years
BTW.......the time required for any investment to double at this interest rate is not dependent upon the size of the original investment.....!!!!
