log (5x-1) - log (X-2) = log 3 solve for x

We'll impose constraints of existence of
logarithms:

5x -
1>0

5x>1

x>1/5

and

x-2>0

x>2

The
common interval of admissible values is
(2,+infinite).

We'll add log (X-2) both
sides:

log (5x-1)  = log 3 + log
(X-2)

log (5x-1)  = log
3*(X-2)

Since the bases are matching, we'll apply one to
one property:

5x - 1 = 3x -
6

We'll subtract 3x:

2x - 1 =
-6

We'll add 1:

2x =
-5

x = -5/2

Since the value
-5/2 doesn't belong to the interval of admissible values, teh equation has no
solution.