What is 1/(2^{log_{10)(x^2-1)}}>1 ?

The solution to 1/(2^{log_{10)(x^2-1)}}>1 is -sqrt(2)<x<-1\lor 1<x<sqrt(2)

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1/(2^{log_{10)(x^2-1)}}>1

\frac{1}{2 ^{l o g_{10} (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_{10} (x^{2} - 1)}} > 1

Let

u = lo g_{10} (x^{2} - 1)

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

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

u < 0

Substitute back

u = lo g_{10} (x^{2} - 1)

lo g_{10} (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_{3} (x^{2} - 16) < 2

lo g_{10} (1 - x) < - 1

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

lo g_{2} (x) < x

lo g_{10} (\frac{( x ^{2} - 36 )}{( 48 + 22 x - 5 x ^{2} )}) \geq 0

What is 1/(2^{log_{10)(x^2-1)}}>1 ?
The solution to 1/(2^{log_{10)(x^2-1)}}>1 is -sqrt(2)<x<-1\lor 1<x<sqrt(2)