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

There is no graph but what i can say is that 1 is the slope and that -2 is the y- intercept