What is log_{2}(x)>= 1+log_{2}(2-x) ?

The solution to log_{2}(x)>= 1+log_{2}(2-x) is 4/3 <= x<2

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log_{2}(x)>= 1+log_{2}(2-x)

lo g_{2} (x) \geq 1 + lo g_{2} (2 - x)

\frac{4}{3} \leq x < 2

Interval Notation

[\frac{4}{3}, 2)

Decimal Notation

1.33333 \dots \leq x < 2

Solution steps

lo g_{2} (x) \geq 1 + lo g_{2} (2 - x)

Apply log rules

x \geq 2 (2 - x) and 2 (2 - x) > 0

x \geq 2 (2 - x) and 2 (2 - x) > 0 : \frac{4}{3} \leq x < 2

\frac{4}{3} \leq x < 2

Find positive values for logs:

0 < x < 2

Combine the intervals

\frac{4}{3} \leq x < 2 and 0 < x < 2

Merge Overlapping Intervals

\frac{4}{3} \leq x < 2

2 x^{2} + 8 x - 6 = 0

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(- 3 + 2 i)^{2}

5 y^{2} + 20 = 0

What is log_{2}(x)>= 1+log_{2}(2-x) ?
The solution to log_{2}(x)>= 1+log_{2}(2-x) is 4/3 <= x<2