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.

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

You just need to add 32 to 14 and get 46.