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If log_(0.3)(x - 1) < log_(0.09)(x - 1), then x lies in which interval?

1. (2, inf)
2. (-2, -1)
3. (1, 2)
4. (0, inf)

Correct answer: (2, inf)

Solution

Since 0.09 = (0.3)², log_(0.09)(x-1) = log_(x-1) / log(0.09) = (1/2) * log_(0.3)(x-1). Let t = log_(0.3)(x-1); the inequality t < t/2 gives t/2 < 0, so t < 0, meaning log_(0.3)(x-1) < 0. Because base 0.3 < 1, this means x - 1 > 1, i.e., x > 2.

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