Question 387079
log x + log(x-1) = log(4x)
log x(x-1) = log(4x)
x(x-1) = (4x)
x^2-x = 4x
x^2-5x = 0
x(x-5) = 0
x = {0,5}
Testing for "extraneous solutions" we must toss out 0 leaving:
x = 5