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

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log (5x-1) - log (x-2) = log 3

First we will simplify the equation:

We know that:

log a = log b = log a/b

==> log (5x-1)/(x-2) = log 3

Also, we know that:

if log a = log b ==> a = b

==> (5x-1) /(x-2) = 3

Now multiply by (x-2):

==> (5x-1) = 3(x-2)

==> 5x-1 = 3x - 6

Combine like terms:

==> 5x-3x = -6 + 1

==> 2x = -5

**==. X= -5/2**

**But the function is not defined when x= -5/2 **

**==> There is no solution.**

The equation log (5x-1) - log (x-2) = log 3 has to be solved for x.

Use the property of logarithm log a - log b = log(a/b). This gives:

log((5x - 1)/(x - 2)) = log 3

Now the terms on both the sides of the equation of which the logarithm are being taken can be equated. This gives:

(5x - 1)/(x - 2) = 3

5x - 1 = 2*(x - 2)

5x - 1 = 2x - 4

5x - 2x = 4 + 1

3x = 5

x = 5/3

The required solution is x = 5/3

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

We use log a- logb = loga/b to solve the equation.

log(5x-1)-log(x-2) = log(5x-1)/(x-2).

Therefore log(5x-1)/(x-2) = log3

Now we take antilogarithms of both sides:

(5x-1)/(x-2) = 3.

Multiply both sides by x-2.

5x-1 = 3(x-2)

5x-1 = 3x-6

5x-3x = -6+1

2x = -5

x = -5/2.

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.