If $x$ satisfies ${log}_{3} (2 x + 1) < {log}_{3} 5$ , then the intervals containing solution set of x is/are :

(- \frac{1}{2}, 0)

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[0, 2)

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[1, 2)

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(2, 3)

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Solution

The correct options are
A $(- \frac{1}{2}, 0)$
B $[0, 2)$
C $[1, 2)$
$l o g_{3} (2 x + 1) < l o g_{3} (5)$

$l o g_{3} (2 x + 1) - l o g_{3} (5) < 0$

$l o g_{3} (\frac{2 x + 1}{5}) < 0$

$0 < \frac{2 x + 1}{5} < 1$

$0 < 2 x + 1 < 5$

$- 1 < 2 x < 4$

$\frac{- 1}{2} < x < 2$
Hence

$x ϵ (\frac{- 1}{2}, 2)$ .

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