Let upstream be positive, and an upstream velocity be positive.

Let C be the velocity of the current, so C is negative.

Let P be the rate of the paddler in still water. P can be either positive or negative depending on which way he’s rowing.

Let s be position, and let s = 0 at the instant the paddler passes the log.

Let T be the time it takes the log to go 2 miles downstream.

The canoeist goes upstream at a rate of P + C for 1 hour traveling P+C mi.

Then he goes downstream at a rate of -P + C for T – 1 hours.

Here’s the equation:

position of the paddler = position of log = -2 [at time T]

(P + C) + (-P + C)(T – 1) = CT = -2

P + C – PT + P + CT – C = -2

P – PT + P + CT = -2

2P – PT = 0

P(2 – T) = 0

So either P is zero or 2 – T = 0

T = 2

CT = -2

C(2) = -2

C = -1, so the current is 1mi/hr

Oddly enough, as long as C = -1, any value of P faster than the current will satisfy the original equation.