If log 03 x -1-log 0 09 x -1- then x lies in the intervalA- -2-1B- 1-2C- NoneD- 2- -

The correct option is D (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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

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)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(- 2, - 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(1, 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Similar questions

Q. If

l o g_{03} (x - 1) < l o g_{009} (x - 1)

, then x lies in the interval

Find 'x' if:

log (x-1) + log (x+1) = log₂1.

\int_{0}^{3} (x^{2} + 1) d x

\lim_{x \to 0} \frac{3 x + 1}{x + 3}

Q. If

y = (x - 1) \log (x - 1) - (x + 1) \log (x + 1)

, prove that

\frac{d y}{d c} = \log (\frac{x - 1}{1 + x})

Join BYJU'S Learning Program

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