Begin with d=rt, where distance is in miles, rate is in miles per hour, and time is in hours. It took the woman 2 hours to paddel 2 miles up stream. Since the effect of her paddling on her overall net speed is constant and the river current rate is constant, we have the equation

2=(r_w-r_c)2 (1)

Solving for r_w yields

r_w=1+r_c (2)

Again, return to d=rt to model her downstream path. She paddled downstream for 2 miles this time making the stream current additive, but did so for an unspecified amount of time. The equation to represent this is

2=(r_w+r_c)t (3)

Substituting (2) for r_w yields

2=(1+r_c+r_c)t

2/t=1+2r_c

2r_c=2/t-1

r_c=1/t-1/2 (4)

Now examine the log. Return to d=rt and model what happens. The log travels 1 mile, but for 1+t hours. 1+t comes from the fact that the log is floating downstream while the woman is still paddling upstream for 1 hour and then paddling downstream for the unspecified amount of time, t. The equation for this is

1=r_c(1+t) (5)

1/r_c=1+t

1/r_c-1=t (6)

Equations (4) and (6) can now be used to solve for r_c. The work is below

1/(1/r_c-1)=r_c+1/2

1=(r_c+1/2)(1/r_c-1)

1=1-r_c+1/(2r_c)-1/2

0=-r_c-1/2+1/(2r_c)

0=-r_c^2-r_c/2+1/2

r_c^2+r_c/2-1/2=0

r_c=-1/4+sqrt(1/4+2)/2

r_c=-1/4+3/4

r_c=1/2

Thus the rate of the current is 1/2 miles per hour, or at least I think so.