What is (log_{10}(0.5x))^2+log_{10}(0.5x)-2<= 0 ?

The solution to (log_{10}(0.5x))^2+log_{10}(0.5x)-2<= 0 is 1/50 <= x<= 20

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(log_{10}(0.5x))^2+log_{10}(0.5x)-2<= 0

(lo g_{10} (0.5 x))^{2} + lo g_{10} (0.5 x) - 2 \leq 0

\frac{1}{50} \leq x \leq 20

Interval Notation

[\frac{1}{50}, 20]

Decimal Notation

0.02 \leq x \leq 20

Solution steps

(lo g_{10} (0.5 x))^{2} + lo g_{10} (0.5 x) - 2 \leq 0

Simplify

lo g_{10} (0.5 x)^{2} + lo g_{10} (0.5 x) - 2 \leq 0

Let

u = lo g_{10} (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_{10} (0.5 x)

- 2 \leq lo g_{10} (0.5 x) \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq lo g_{10} (0.5 x) and lo g_{10} (0.5 x) \leq 1

- 2 \leq lo g_{10} (0.5 x) : x \geq \frac{1}{50}

lo g_{10} (0.5 x) \leq 1 : 0 < x \leq 20

Combine the intervals

x \geq \frac{1}{50} and 0 < x \leq 20

Merge Overlapping Intervals

\frac{1}{50} \leq x \leq 20

lo g_{\frac{1}{2}} (x^{2} - 1) \geq lo g_{\frac{1}{2}} (3 x)

3 x + 1 < 2 \cdot lo g_{2} (x + 4)

lo g_{10} (x - 2) (3 x^{2} - x - 1) > 0

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

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

What is (log_{10}(0.5x))^2+log_{10}(0.5x)-2<= 0 ?
The solution to (log_{10}(0.5x))^2+log_{10}(0.5x)-2<= 0 is 1/50 <= x<= 20