lg(5x + 2) = lg(2x-1)

We know that:

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

==> 5x + 2 = 2x -1

Group similar:

==> 3x = -3

==> x= -1

Let us substitute to check answer:

lg(5x+2) = log(2x-1)

lg(-3) = lg(-3)

It is true but impossible, the function is not defined for x=-1

Then the equality has no **real **solution.

First, we'll have to impose constraints of existance of logarithms.

The first constraint: 5x+2>0

We'll subtract 2 both sides:

5x>-2

We'll divide by 5 both sides:

x>-2/5

The second constraint: 2x-1>0

We'll add 1 both sides:

2x>1

We'll divide by 2 both sides:

x>1/2

The values of x which satisfy both constraints are: (1/2 , +inf.)

Now, we'll solve the equation, using the one to one property of logarithms:

5x+2 = 2x-1

We'll subtract 2x both sides:

5x - 2x + 2 = -1

3x + 2 = -1

We'll subtract 2 both sides:

3x = -3

We'll divide by 3:

x = -1

But the calculated value is negative, so the conclusion is: the identity is not true for any value of the variable x!