Solve log2x 1 <= logx 4

Question

Solve the inequality

x \geq 32

Alternative Form

x \in [32, + \infty)

Evaluate

lo g_{10} (2 x) \times 1 \leq lo g_{10} (x) \times 4

Find the domain

More Steps

lo g_{10} (2 x) \times 1 \leq lo g_{10} (x) \times 4, x > 0

Multiply the terms

lo g_{10} (2 x) \leq lo g_{10} (x) \times 4

Multiply the terms

lo g_{10} (2 x) \leq 4 lo g_{10} (x)

Move the expression to the left side

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

Add the terms

More Steps

lo g_{10} (\frac{2}{x ^{3}}) \leq 0

For 10 >1 the expression lo g_{10} (\frac{2}{x ^{3}}) \leq 0 is equivalent to \frac{2}{x ^{3}} \leq 1 0^{0}

\frac{2}{x ^{3}} \leq 1 0^{0}

Evaluate the power

\frac{2}{x ^{3}} \leq 1

Calculate

\frac{2}{x ^{3}} - 1 \leq 0

Calculate

More Steps

\frac{2 - x ^{3}}{x ^{3}} \leq 0

Separate the inequality into 2 possible cases

{2 - x^{3} \geq 0 x^{3} < 0 {2 - x^{3} \leq 0 x^{3} > 0

Solve the inequality

More Steps

{x \leq 32 x^{3} < 0 {2 - x^{3} \leq 0 x^{3} > 0

The only way a base raised to an odd power can be less than 0 is if the base is less than 0

{x \leq 32 x < 0 {2 - x^{3} \leq 0 x^{3} > 0

Solve the inequality

More Steps

{x \leq 32 x < 0 {x \geq 32 x^{3} > 0

The only way a base raised to an odd power can be greater than 0 is if the base is greater than 0

{x \leq 32 x < 0 {x \geq 32 x > 0

Find the intersection

x < 0 {x \geq 32 x > 0

Find the intersection

x < 0 x \geq 32

Find the union

x \in (- \infty, 0) \cup [32, + \infty)

Check if the solution is in the defined range

x \in (- \infty, 0) \cup [32, + \infty), x > 0

Solution

x \geq 32

Alternative Form

x \in [32, + \infty)

Show Solution