We have to use the equation lg x / (1 - lg 2) = 2 to find the solution in the simplest form.

Now lg x / (1- lg 2) = 2

=> lg x = 2*(1- lg 2)

=> lg x = 2 - 2 lg 2

=> lg x + lg 2^2 = 2

=> lg ( 4x) = 2

assuming lg is the logarithm to the base 10, we raise 10 to the power of the two sides of the equation.

=> 10^( lg 4x) = 10^2

=> 4x = 100

=> x = 25.

**Therefore the required value of x is 25.**

(lg x) / (1 - lg 2) = 2

lg x = 2 - 2 lg 2

RHS (Right Hand Side)

= 2 - 2 lg 2

= 2 lg 10 - 2 lg 2

= lg 10^2 - lg 2^2

= lg (100/4)

= lg 25

Comparing the argument of lg for LHS and RHS,

**x = 25**

We'll impose the constraints of existence of logarithm.

x>0

The solution has to be in the interval of admissible values (0,+infinite)

lgx/(1-lg2) = 2

lgx = 2 - 2*lg2

Well use the power rule of logarithms for 2*lg2:

a*lg b = lg b^a

2*lg2 = lg 2^2 = lg 4

lgx = 2 - lg 4

But 2 = 2*1 = 2lg 10 = lg 10^2 = lg 100

We'll re-write the equation:

lgx = lg 100 - lg 4

We'll use the quotient rule of logarithms:

lg 100 - lg 4 = lg 100/4

lg 100 - lg 4 = lg 25

lgx = lg 25

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

**x = 25**

**Since the solution belongs to the interval of admissible value, we'll accept it.**