If log 03 x -1-log 0 09 x -1- then x lies in the intervalA- 2- -B- -2-1C- 1-2D- None

The correct option is A (2, ∞ )log_03 (x 1) lt;log_(03)^2(x 1)=1/2log_03(x 1) ∴ 1/2log_03(x 1) lt;0 or log_03 (x 1) lt;0=log 1 or (x 1) gt;1 or x g ...

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Byju's Answer

Standard X

Mathematics

Logarithmic inequalities

If log 03 x -...

Question

If $l o g_{03} (x - 1) < l o g_{009} (x - 1)$ , then x lies in the interval

(2, \infty)

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(- 2, - 1)

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(1, 2)

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None

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Solution

The correct option is A $(2, \infty)$
$l o g_{03} (x - 1) < l o g_{(03)^{2}} (x - 1) = \frac{1}{2} l o g_{03} (x - 1)$
$∴ \frac{1}{2} l o g_{03} (x - 1) < 0$
or $l o g_{03} (x - 1) < 0 = l o g 1$
or $(x - 1) > 1 o r x > 2$
as base is less than 1, therefore the inequality is reversed, now $x > 2 \Rightarrow x$ lies in $(2, \infty)$

Suggest Corrections

If log03(x−1)<log0 09(x−1), then x lies in the interval

The correct option is A (2,∞)log03 (x−1)<log(03)2(x−1)=12log03(x−1) ∴ 12log03(x−1)<0 or log03 (x−1)<0=log 1 or (x−1)>1 or x>2 as base is less than 1, therefore the inequality is reversed, now x>2⇒x lies in (2,∞)

If $l o g_{03} (x - 1) < l o g_{009} (x - 1)$ , then x lies in the interval