The inequality \(\log_{a}{f(x)} < \log_{a}{g(x)}\) implies that

Question and options

The inequality

\log_{a}{f(x)} < \log_{a}{g(x)}

implies that

$f(x) > g(x) > 0$ for $0 < a < 1$ and $g(x) > f(x) > 0$ for $a > 1$
$g(x) > f(x) > 0$ for $0 < a < 1$ and $f(x) > g(x) > 0$ for $a > 1$
$f(x) > g(x) > 0$ for $a > 0$
$g(x) > f(x) > 0$ for $a > 0$

Answer and explanation

Correct answer: A - \(f(x) > g(x) > 0\) for \(0 < a < 1\) and \(g(x) > f(x) > 0\) for \(a > 1\)

Logs need positive arguments; log_a is decreasing for 0<a<1 (so f > g > 0) and increasing for a > 1 (so g > f > 0).

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