We have to solve the equation: log(x) 2 + log(2x) 2 = log(4x) 2

Now log(a) b = 1/ log (b) a

So we can write

log(x) 2 + log(2x) 2 = log(4x) 2

=> 1/log (2) x + 1/ log 2( 2x) = 1/ log 2 ( 4x)

=>...

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We have to solve the equation: log(x) 2 + log(2x) 2 = log(4x) 2

Now log(a) b = 1/ log (b) a

So we can write

log(x) 2 + log(2x) 2 = log(4x) 2

=> 1/log (2) x + 1/ log 2( 2x) = 1/ log 2 ( 4x)

=> 1/ log (2) x + 1/[ log (2) x + log 2 (2)] = 1/ [ log(2)x + log(2) 2^2]

=> 1/ log(2)x + 1/[log(2)x + 1] = 1/[log(2)x + 2]

Let log(2)x = y

=> 1/y + 1/(y +1) = 1/(y+2)

=> [y+1 +y]/y*(y+1) = 1/(y+2)

=> (2y+1)(y+2) = y(y+1)

=> 2y^2 + 4y +y +2 = y^2 + y

=> y^2 + 4y + 2 = 0

y1 = [-4 + sqrt (16 - 8)]/2

=> [-2 + sqrt 2]

y2 = -2 - sqrt 2

y = log (2) x

log(2)x = -2 + sqrt 2

=> x = 2^(-2 + sqrt 2)

and x = 2^(-2 - sqrt 2)

**Therefore x = 2^(-2 + sqrt 2) and 2^(-2 - sqrt 2)**