What is log_{1/2}(log_{8}((x^2-2x)/(x-3)))>0 ?

The solution to log_{1/2}(log_{8}((x^2-2x)/(x-3)))>0 is 4<x<6

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log_{1/2}(log_{8}((x^2-2x)/(x-3)))>0

lo g_{\frac{1}{2}} (lo g_{8} (\frac{x ^{2} - 2 x}{x - 3})) > 0

4 < x < 6

Interval Notation

(4, 6)

Solution steps

lo g_{\frac{1}{2}} (lo g_{8} (\frac{x ^{2} - 2 x}{x - 3})) > 0

Apply log rules

lo g_{8} (\frac{x ^{2} - 2 x}{x - 3}) < 1 and lo g_{8} (\frac{x ^{2} - 2 x}{x - 3}) > 0

lo g_{8} (\frac{x ^{2} - 2 x}{x - 3}) < 1 : 0 < x < 2 or 4 < x < 6

lo g_{8} (\frac{x ^{2} - 2 x}{x - 3}) > 0 : x > 3

Combine the intervals

(0 < x < 2 or 4 < x < 6) and x > 3

Merge Overlapping Intervals

4 < x < 6

36 = q^{2}

7 a^{2} b + ab

7 x - 4 (x - 5) - 8 x < 2 (x + 5)

4 x^{2} - 3 x + 1 = 0

15 x^{2} + 58 x + 48 = 0

What is log_{1/2}(log_{8}((x^2-2x)/(x-3)))>0 ?
The solution to log_{1/2}(log_{8}((x^2-2x)/(x-3)))>0 is 4<x<6