Question

jocelyn invests $1200 in an account that earns 2.4% annual interest. Marcus invests $400 in an account that earns 5.2% annual interest. Find when the value of Marcus's investment equals the value of jocelyn's investment and find the common value of the investments at that time.

Answer 1

Let the time that the two investments be n, then the value of the investment at time n is given by P(1 + r)^n where P is the invested amount, r is the rate.
Thus, 1200(1 + 2.4/100)^n = 400(1 + 5.2)^n
1200(1 + 0.024)^n = 400(1 + 0.052)^n
1200(1.024)^n = 400(1.052)^n
1200 / 400 = (1.052)^n / (1.024)^n
3 = (1.052 / 1.024)^n = (1.02734375)^n
log 3 = log (1.02734375)^n = n log (1.02734375)
n = log 3 / log (1.02734375) = 40.72 years.
Therefore, the two investments becomes equal after 40.72 years.

The common value of the investment after 40.72 years is 1200(1.024)^40.72 = 1200 x 2.627 = $3,152.42