Solve (log3(3x^25) - log3^(1/2)(x 7))/(x^4 -81) >= 0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, - 3) \cup [0, \frac{7}{30}] \cup (3, + \infty)

Evaluate

\frac{lo g _{10} ( 3 ) \times ( 3 x ^{2} \times 5 ) - lo g _{10} ( 3 ^{\frac{1}{2}} ) \times ( x \times 7 )}{x ^{4} - 81} \geq 0

Find the domain

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\frac{lo g _{10} ( 3 ) \times ( 3 x ^{2} \times 5 ) - lo g _{10} ( 3 ^{\frac{1}{2}} ) \times ( x \times 7 )}{x ^{4} - 81} \geq 0, x \in (- \infty, - 3) \cup (- 3, 3) \cup (3, + \infty)

Remove the parentheses

\frac{lo g _{10} ( 3 ) \times 3 x ^{2} \times 5 - lo g _{10} ( 3 ^{\frac{1}{2}} ) \times x \times 7}{x ^{4} - 81} \geq 0

Simplify

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\frac{30 lo g _{10} ( 3 ) \times x ^{2} - 7 lo g _{10} ( 3 ) \times x}{2 ( x ^{4} - 81 )} \geq 0

Set the numerator and denominator of \frac{30 lo g _{10} ( 3 ) \times x ^{2} - 7 lo g _{10} ( 3 ) \times x}{2 ( x ^{4} - 81 )} equal to 0 to find the values of x where sign changes may occur

30 lo g_{10} (3) \times x^{2} - 7 lo g_{10} (3) \times x = 0 2 (x^{4} - 81) = 0

Calculate

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x = 0 x = \frac{7}{30} 2 (x^{4} - 81) = 0

Calculate

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x = 0 x = \frac{7}{30} x = 3 x = - 3

Determine the test intervals using the critical values

x < - 3 - 3 < x < 0 0 < x < \frac{7}{30} \frac{7}{30} < x < 3 x > 3

Choose a value form each interval

x_{1} = - 4 x_{2} = - 2 x_{3} = \frac{7}{60} x_{4} = 2 x_{5} = 4

To determine if x < - 3 is the solution to the inequality,test if the chosen value x = - 4 satisfies the initial inequality

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x < - 3 is the solution x_{2} = - 2 x_{3} = \frac{7}{60} x_{4} = 2 x_{5} = 4

To determine if - 3 < x < 0 is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

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x < - 3 is the solution - 3 < x < 0 is not a solution x_{3} = \frac{7}{60} x_{4} = 2 x_{5} = 4

To determine if 0 < x < \frac{7}{30} is the solution to the inequality,test if the chosen value x = \frac{7}{60} satisfies the initial inequality

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x < - 3 is the solution - 3 < x < 0 is not a solution 0 < x < \frac{7}{30} is the solution x_{4} = 2 x_{5} = 4

To determine if \frac{7}{30} < x < 3 is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

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x < - 3 is the solution - 3 < x < 0 is not a solution 0 < x < \frac{7}{30} is the solution \frac{7}{30} < x < 3 is not a solution x_{5} = 4

To determine if x > 3 is the solution to the inequality,test if the chosen value x = 4 satisfies the initial inequality

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x < - 3 is the solution - 3 < x < 0 is not a solution 0 < x < \frac{7}{30} is the solution \frac{7}{30} < x < 3 is not a solution x > 3 is the solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x < - 3 is the solution 0 \leq x \leq \frac{7}{30} is the solution x > 3 is the solution

The final solution of the original inequality is x \in (- \infty, - 3) \cup [0, \frac{7}{30}] \cup (3, + \infty)

x \in (- \infty, - 3) \cup [0, \frac{7}{30}] \cup (3, + \infty)

Check if the solution is in the defined range

x \in (- \infty, - 3) \cup [0, \frac{7}{30}] \cup (3, + \infty), x \in (- \infty, - 3) \cup (- 3, 3) \cup (3, + \infty)

Solution

x \in (- \infty, - 3) \cup [0, \frac{7}{30}] \cup (3, + \infty)

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