# Solve log(x)2 . log(2x)4 > 1 for x where log(a)b denotes log of b to the base a.

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We'll use the formula that is interchanging the base and argument.

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

According to this formula, we'll have:

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

We'll apply the product rule of logarithms:

1/log(4) 2x =1/[log(4) 2 + log(4) x]

1/log(4) 2x =1/[1/log(2) 4 + 1/log(x) 4]

1/log(4) 2x =1/[1/2log(2) 2 + 1/2log(x) 2]

1/log(4) 2x =1/[1/2 + 1/2log(x) 2]

1/log(4) 2x =2log(x) 2/[log(x) 2 + 1]

But 1/log(4) 2x > 1 => 2log(x) 2/[log(x) 2 + 1] > 1

2log(x) 2/[log(x) 2 + 1] - 1 > 0

[2log(x) 2 - log(x) 2 - 1]/[log(x) 2 + 1] - 1 > 0

[log(x) 2 - 1]/[log(x) 2 + 1] - 1 > 0

The fraction is positive if both, numerator and denominator are positive.

log(x) 2 - 1 > 0

log(x) 2 > 1

2 > x

log(x) 2 + 1 > 0

log(x) 2 > -1

2 > 1/x

2x > 1

x > 1/2

**The values of x for log(2x)4 > 1 belong to the intervals (1/2 ; 2).**