Final answer:
To solve 0 < LOG x²+2x-2 < 1, we translate the inequalities into exponentiated form and solve for the intervals that satisfy both. After solving each quadratic inequality separately, we can determine the interval of x by finding the intersection of the intervals obtained.
Step-by-step explanation:
We want to solve the inequality 0 < LOG x²+2x-2 < 1 for the interval of values of x. First, we interpret the inequality in terms of the natural logarithm since LOG usually refers to the base 10 logarithm, but without further context, we'll assume it's base e.
By exponentiating each part of the inequality, we first solve for e⁰ < x² + 2x - 2 < e¹. This gives us the inequalities x² + 2x - 2 > 1 and x² + 2x - 2 < e.
To find the intervals, we need to solve the two quadratic inequalities. For the first inequality, we are looking for solutions to x² + 2x - 3 > 0. By factoring, we get (x+3)(x-1) > 0, which implies the solutions are x < -3 or x > 1. For the second inequality, we are looking for solutions to x² + 2x - (e + 2) < 0. Completing the square or using the quadratic formula will give us the interval that satisfies this inequality.
Without the exact numerical bounds provided by the solutions to these inequalities, which would require further calculation, we cannot specify the exact intervals. However, once found, the final solution is where the intervals from each inequality overlap.