What is (log_{2}(x))^2<2log_{2}(x)+3 ?

The solution to (log_{2}(x))^2<2log_{2}(x)+3 is 1/2 <x<8

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(log_{2}(x))^2<2log_{2}(x)+3

(lo g_{2} (x))^{2} < 2 lo g_{2} (x) + 3

\frac{1}{2} < x < 8

Interval Notation

(\frac{1}{2}, 8)

Decimal Notation

0.5 < x < 8

Solution steps

(lo g_{2} (x))^{2} < 2 lo g_{2} (x) + 3

Simplify

lo g_{2} (x)^{2} < 2 lo g_{2} (x) + 3

Move

2 lo g_{2} (x)

to the left side

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

Let

u = lo g_{2} (x)

u^{2} - 2 u < 3

u^{2} - 2 u < 3 : - 1 < u < 3

- 1 < u < 3

Substitute back

u = lo g_{2} (x)

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

a < u < b

then

a < u and u < b

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

- 1 < lo g_{2} (x) : x > \frac{1}{2}

lo g_{2} (x) < 3 : 0 < x < 8

Combine the intervals

x > \frac{1}{2} and 0 < x < 8

Merge Overlapping Intervals

\frac{1}{2} < x < 8

6 > p

750

∣ 2 x + 4 ∣ < 7

\frac{3 x}{7} = \frac{1}{2}

What is (log_{2}(x))^2<2log_{2}(x)+3 ?
The solution to (log_{2}(x))^2<2log_{2}(x)+3 is 1/2 <x<8