Solve log2(4x) log2(3)<log2(6x^24)

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

Solve for x

x \in (- \infty, 0) \cup (\frac{lo g _{10} ( 2 )}{2}, + \infty)

Evaluate

lo g_{10} (2) \times 4 x lo g_{10} (2) \times 3 < lo g_{10} (2) \times (6 x^{2} \times 4)

Remove the parentheses

lo g_{10} (2) \times 4 x lo g_{10} (2) \times 3 < lo g_{10} (2) \times 6 x^{2} \times 4

Multiply

More Steps

12 (lo g_{10} (2))^{2} x < lo g_{10} (2) \times 6 x^{2} \times 4

Multiply

More Steps

12 (lo g_{10} (2))^{2} x < 24 lo g_{10} (2) \times x^{2}

Move the expression to the left side

12 (lo g_{10} (2))^{2} x - 24 lo g_{10} (2) \times x^{2} < 0

Rewrite the expression

12 (lo g_{10} (2))^{2} x - 24 lo g_{10} (2) \times x^{2} = 0

Factor the expression

More Steps

12 lo g_{10} (2) \times x (lo g_{10} (2) - 2 x) = 0

When the product of factors equals 0,at least one factor is 0

12 lo g_{10} (2) \times x = 0 lo g_{10} (2) - 2 x = 0

Solve the equation for x

x = 0 lo g_{10} (2) - 2 x = 0

Solve the equation for x

More Steps

x = 0 x = \frac{lo g _{10} ( 2 )}{2}

Determine the test intervals using the critical values

x < 0 0 < x < \frac{lo g _{10} ( 2 )}{2} x > \frac{lo g _{10} ( 2 )}{2}

Choose a value form each interval

x_{1} = - 1 x_{2} = \frac{lo g _{10} ( 2 )}{4} x_{3} = 1

To determine if x < 0 is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < 0 is the solution x_{2} = \frac{lo g _{10} ( 2 )}{4} x_{3} = 1

To determine if 0 < x < \frac{lo g _{10} ( 2 )}{2} is the solution to the inequality,test if the chosen value x = \frac{lo g _{10} ( 2 )}{4} satisfies the initial inequality

More Steps

x < 0 is the solution 0 < x < \frac{lo g _{10} ( 2 )}{2} is not a solution x_{3} = 1

To determine if x > \frac{lo g _{10} ( 2 )}{2} is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

More Steps

x < 0 is the solution 0 < x < \frac{lo g _{10} ( 2 )}{2} is not a solution x > \frac{lo g _{10} ( 2 )}{2} is the solution

Solution

x \in (- \infty, 0) \cup (\frac{lo g _{10} ( 2 )}{2}, + \infty)

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