lg (2x - sqrt2) = lg[1/(2x+sqrt2)]

We know that if log a = log b , then a = b

==> *2x - sqrt2 = 1/(2x+ sqrt2)

Now cross multiply:

(2x - sqrt2)*(2x+ sqrt2) = 1

Open brackets:

4x^2 - 2 = 1

==> 4x^2 = 3

==> x^2 = 3/4

==> x1 = -sqrt3/2 ( impossible)

==> **x2= sqrt3/ 2**

First, we'll impose the constraints of existence of logarithms:

2x+sqrt2>0

2x>-sqrt2

We'll divide by 2:

x>-sqrt2/2

2x-sqrt2>0

2x>sqrt2

We'll divide by 2:

x>sqrt2/2

The common interval of admissible values for x is (sqrt2/2,+inf.).

Because the bases of logarithms are matching, we'll apply the one to one property of logarithms:

(2x-sqrt2) = [1/(2x+sqrt2)]

We'll cross multiply:

(2x-sqrt2)*(2x+sqrt2) = 1

We transform the product into a difference of squares:

4x^2 - 2 =1

We'll add 2 both sides:

4x^2 = 1+2

4x^2 = 3

We'll divide by 4:

x^2 = 3/4

**x1 = +sqrt3/2**

**x2 = -sqrt3/2 **

**Since -sqrt3/2 doesn't belong to the interval of admissible values, the only solution of the equation is x1 = +sqrt3/2.**

To solve lg(2x-sqrt2) = lg {1/(2x+sqrt2)}.

Take antilogarithms .

2x-sqrt2 = 1/(2x+sqrt2). Cross multiply.

(2x-sqrt2)(2x+sqrt2) = 1

4x^2 - 2 = 1

4x^2 = = 1+2 =3

4x^2 = 3

x^2 = 3/4

x = sqrt(3/4) , or - sqrt(3/4)

x = (1/2)sqrt3 , or - (1/2) sqrt(3)