If log32,log3(2x−5) and log3(2x−72) are in AP,x is equal to

∵log32,log3(2x−5) and log3(2x−72) are in AP ....(i)For defined, 2x−5>0 and 2x−72>0 ∴ 2x>5 ....(ii)From Eq. (i), 2,2x−5,2x−72 are in GP ∴(2x−5)2=2·(2x−72)⇒22x−12·2x+32=0⇒(2x−8)(2x−4)=0∴2x=8,4⇒2x=8=23,2x≠4 [fromEq.(ii)]∴x=3

If log32,log3(2x−5) and log3(2x−72) are in AP,x is equal to

∵log32,log3(2x−5) and log3(2x−72) are in AP ....(i)For defined, 2x−5>0 and 2x−72>0 ∴ 2x>5 ....(ii)From Eq. (i), 2,2x−5,2x−72 are in GP ∴(2x−5)2=2·(2x−72)⇒22x−12·2x+32=0⇒(2x−8)(2x−4)=0∴2x=8,4⇒2x=8=23,2x≠4 [fromEq.(ii)]∴x=3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\log_{3} 2, \log_{3} (2^{x} - 5)$ and $\log_{3} (2^{x} - \frac{7}{2})$ are in $AP, x$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2,3

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$∵ \log_{3} 2, \log_{3} (2^{x} - 5)$ and $\log_{3} (2^{x} - \frac{7}{2})$ are in AP ....(i)
For defined, $2^{x} - 5 > 0$ and $2^{x} - \frac{7}{2} > 0$
$∴ 2^{x} > 5$ ....(ii)
From Eq. (i), $2, 2^{x} - 5, 2^{x} - \frac{7}{2}$ are in GP
$\begin{array}{l} ∴ & {(2^{x} - 5)}^{2} = 2 \cdot (2^{x} - \frac{7}{2}) \\ \Rightarrow & 2^{2 x} - 12 \cdot 2^{x} + 32 = 0 \\ \Rightarrow & (2^{x} - 8) (2^{x} - 4) = 0 \\ ∴ & 2^{x} = 8, 4 \\ \Rightarrow & 2^{x} = 8 = 2^{3}, 2^{x} \neq 4 [fromEq . (ii)] \\ ∴ & x = 3 \end{array}$