wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If log03(x1)<log0 09(x1), then x lies in the interval

A
(2,)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(2,1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(1,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (2,)
log03 (x1)<log(03)2(x1)=12log03(x1)
12log03(x1)<0
or log03 (x1)<0=log 1
or (x1)>1 or x>2
as base is less than 1, therefore the inequality is reversed, now x>2x lies in (2,)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Logarithmic Inequalities
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon