I remember asking you this question last year in the marathon thread. Or maybe you asked me?Let the volume be T
T/80 = rate of both pipes
Let x be time in minutes
T/(x-120) + T/x = T/80 since it's both pipes working together
2x-120 = x(x-120)/80. Solve for x. x will be the larger solution as one of them will give a negative time value
I didn't considered that, I only tried the odd = even contradiction, thanks Jm.For the other one...
Suppose that log_3 5 = p/q p,q relatively prime
5^q=3^p, which is impossible if p and q are relatively prime (5^q ends with 5 or 0 and 3^p ends with 3, 9, 7, 1 or 3)