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Solve for the interval of values of X for which 0<_ LOG x2 2x-2<_ 1

User Mikeware
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The interval of values of x for which 0 < log(x^2 - 2x - 2) < 1 is:

x ∈ (-1, 1 - √3) ∪ (1 + √3, 3)

The interval of values of x for which 0 < log(x^2 - 2x - 2) < 1:

Step 1: Rewrite the inequalities.

Since the logarithm is only defined for positive values, we know that x^2 - 2x - 2 > 0. Additionally, the inequality 0 < log(x^2 - 2x - 2) < 1 implies that 1 < x^2 - 2x - 2 < 10.

Step 2: Solve the quadratic equation.

The inequality x^2 - 2x - 2 > 0 can be solved using the quadratic formula

x = (2 ± √(2^2 + 4 * 2 * 2)) / 2 = (2 ± √12) / 2 = 1 ± √3

Therefore, x^2 - 2x - 2 > 0 for x < 1 - √3 or x > 1 + √3.

Step 3: Solve the other inequality.

The inequality 1 < x^2 - 2x - 2 < 10 can be rewritten as:

0 < x^2 - 2x - 3 < 9

(x - 3)(x + 1) < 0

This inequality is true when x is between -1 and 3, exclusive.

Step 4: Combine the solutions.

Combining the solutions from steps 2 and 3, we have:

x ∈ (-∞, 1 - √3) ∪ (1 + √3, ∞) ∩ (-1, 3)

Therefore, the interval of values of x for which 0 < log(x^2 - 2x - 2) < 1 is:

x ∈ (-1, 1 - √3) ∪ (1 + √3, 3)

User Jamison Dance
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