CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

Find the value of x which satisfy following equation log0.3(x2x+1)>0.

A
0<x<1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1<x<3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2<x<4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 0<x<1
log0.3(x2x+1)>0.
x2x+1<(.3)0=1
x2x<0
x(x1)<0
0<x<1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Fundamental Laws of Logarithms
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon