You need to solve for x the given equation, such that:

`2lg x = lg(5x - 4)`

Using the power property of logarithms to the left side term, yields:

`lg x^2 = lg (5x - 4)`

Equating the arguments of logarithms yields:

`x^2 = 5x - 4`

You need to move all terms to one side such that:

`x^2 - 5x + 4 = 0`

Using quadratic formula yields:

`x_(1,2) = (5+-sqrt(25 - 16))/2 => x_(1,2) = (5+-3)/2 => x_1 = 4; x_2 = 1`

You need to test the values `x_1 = 4; x_2 = 1` in equation, such that:

`x = 1 => 2lg 1/lg(5 - 4) = 1 => 2*0/0 = 1` invalid (the division by 0 is invalid)

`x = 4 => 2lg 4/lg(20 - 4) = 1 => 2lg 4/lg 16 =1 => lg 4^2/lg 16 = 1` valid

**Hence, evaluating the solutions to the given equation, rejecting the invalid value `x = 1` , yields `x = 4` .**

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