What is (log_{0.5}(x))^2+log_{0.5}(x)-2<= 0 ?

The solution to (log_{0.5}(x))^2+log_{0.5}(x)-2<= 0 is 0.5<= x<= 4

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(log_{0.5}(x))^2+log_{0.5}(x)-2<= 0

(lo g_{0.5} (x))^{2} + lo g_{0.5} (x) - 2 \leq 0

0.5 \leq x \leq 4

Interval Notation

[0.5, 4]

Solution steps

(lo g_{0.5} (x))^{2} + lo g_{0.5} (x) - 2 \leq 0

Simplify

lo g_{0.5} (x)^{2} + lo g_{0.5} (x) - 2 \leq 0

Let

u = lo g_{0.5} (x)

u^{2} + u - 2 \leq 0

u^{2} + u - 2 \leq 0 : - 2 \leq u \leq 1

- 2 \leq u \leq 1

Substitute back

u = lo g_{0.5} (x)

- 2 \leq lo g_{0.5} (x) \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq lo g_{0.5} (x) and lo g_{0.5} (x) \leq 1

- 2 \leq lo g_{0.5} (x) : 0 < x \leq \frac{1}{0.25}

lo g_{0.5} (x) \leq 1 : x \geq 0.5

Combine the intervals

0 < x \leq \frac{1}{0.25} and x \geq 0.5

Merge Overlapping Intervals

0.5 \leq x \leq \frac{1}{0.25}

0.5 \leq x \leq \frac{1}{0.25} : 0.5 \leq x \leq 4

0.5 \leq x \leq 4

2 - lo g_{10} (4 x) > 0

lo g_{x}^{2} (2 + x) < 1

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

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

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

What is (log_{0.5}(x))^2+log_{0.5}(x)-2<= 0 ?
The solution to (log_{0.5}(x))^2+log_{0.5}(x)-2<= 0 is 0.5<= x<= 4