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

The solution to (1/2)^{log_{2}(x^2-1)}>1 is -sqrt(2)<x<-1\lor 1<x<sqrt(2)

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

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

- 2 < x < - 1 or 1 < x < 2

Interval Notation

(- 2, - 1) \cup (1, 2)

Decimal Notation

- 1.41421 \dots < x < - 1 or 1 < x < 1.41421 \dots

Solution steps

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

Let

u = lo g_{2} (x^{2} - 1)

(\frac{1}{2})^{u} > 1

(\frac{1}{2})^{u} > 1 : u < 0

u < 0

Substitute back

u = lo g_{2} (x^{2} - 1)

lo g_{2} (x^{2} - 1) < 0

Apply log rules

x^{2} - 1 < 1 and x^{2} - 1 > 0

x^{2} - 1 < 1 and x^{2} - 1 > 0 : - 2 < x < - 1 or 1 < x < 2

- 2 < x < - 1 or 1 < x < 2

lo g_{4} (x) > 0

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

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

\frac{1}{2 lo g _{10} ( n ^{2} - 1 )} > 1

lo g_{\frac{1}{3}} (3 x + 4) > lo g_{\frac{1}{3}} (x^{2} + 2)

What is (1/2)^{log_{2}(x^2-1)}>1 ?
The solution to (1/2)^{log_{2}(x^2-1)}>1 is -sqrt(2)<x<-1\lor 1<x<sqrt(2)