What is log_{2}(x^2-x)>1-log_{0.5}(x) ?

The solution to log_{2}(x^2-x)>1-log_{0.5}(x) is x>3

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log_{2}(x^2-x)>1-log_{0.5}(x)

lo g_{2} (x^{2} - x) > 1 - lo g_{0.5} (x)

x > 3

Interval Notation

(3, \infty)

Solution steps

lo g_{2} (x^{2} - x) > 1 - lo g_{0.5} (x)

Move

lo g_{0.5} (x)

to the left side

lo g_{2} (x^{2} - x) + lo g_{0.5} (x) > 1

Apply log rules

\frac{x ^{2} - x}{x} > 2

\frac{x ^{2} - x}{x} > 2 : x > 3

x > 3

Find positive values for logs:

x > 1

Combine the intervals

x > 3 and x > 1

Merge Overlapping Intervals

x > 3

\frac{( 4 ^{3} \cdot 4 ^{3} )}{4 ^{3}}

- 3 (\frac{3}{4})^{2}

31 - x = 43

\frac{1}{2 - 1}

What is log_{2}(x^2-x)>1-log_{0.5}(x) ?
The solution to log_{2}(x^2-x)>1-log_{0.5}(x) is x>3