What is log_{4}(x^2+2x)>log_{2}(x-4) ?

The solution to log_{4}(x^2+2x)>log_{2}(x-4) is x>4

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log_{4}(x^2+2x)>log_{2}(x-4)

lo g_{4} (x^{2} + 2 x) > lo g_{2} (x - 4)

x > 4

Interval Notation

(4, \infty)

Solution steps

lo g_{4} (x^{2} + 2 x) > lo g_{2} (x - 4)

Apply log rules

x^{2} + 2 x > x - 4 and x - 4 > 0

x^{2} + 2 x > x - 4 and x - 4 > 0 : x > 4

x > 4

lo g_{\frac{1}{4}} (2 - x) > lo g_{\frac{1}{4}} (\frac{2}{( x + 1 )})

lo g_{3} (x - 2) < 1

(\frac{1}{2})^{l o g_{2} (x^{2} - 1)} > 1

lo g_{4} (x) > 0

lo g_{3 x + 5} (9 x^{2} + 8 x + 8) > 2

What is log_{4}(x^2+2x)>log_{2}(x-4) ?
The solution to log_{4}(x^2+2x)>log_{2}(x-4) is x>4