Question

Solve this application using logarithms.

At what interest rate (to the nearest hundredth of a percent) compounded annually will money in savings double in five years?

Answer 1

(n) = I(o)(1 + r/100)^n is the formula for compound interest.

1).
2 = 1(1 + r/100)^5
2 = (1 + r/100)^5
2^(1/5) = 1 + r/100)
0.2log(10)2 = (100 + r) / 100 = log(10)(100 + r ) - log(10)100
0.060206 = log(10)(!00 + r) - 2
2.060206 = log(10)(100 + r)
Antilog
. 114.8698 = 100 + r
14.6898 = r
r = 14.87%

Answer 2

Medal and fan?! Thx