Len invests $5200 at 3%/a, while his friend Dave invests $3600 at 5%/a. How long will it take for Dave's

investment to be worth the same amount as Len's?

Suppose that it will take n years for Dave's investment to be equal to Len's;

thus using the compound interest formula we shall have:

A=p (1+r/100) ^n

thus the investment for Len after n year will be:

A=5200 (1+3/100) ^n

A=5200 (1.03) ^n

The total amount Dave's amount after n years will be:

A=3600 (1+5/100) ^n

A=3600 (1.05) ^n

since after n years the investments will be equal, the value of n will be calculated as follows;

5200 (1.03) ^n=3600 (1.05) ^n

5200/3600 (1.03) ^n = (1.05) ^n

13/9 (1.03) ^n = (1.05) ^n

introducing the natural logs we get:

ln (13/9) + n ln1.03=n ln 1.05

ln (13/9) = n ln 1.05-n ln 1.03

ln (13/9) = 0.0192n

n=[ln (13/9) ]/[0.0192]

n=19.12

thus the amount will be equal after 19 years