1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Properties of Iota
Number of sol...
Question
Number of solutions satisfying,
√
5
−
l
o
g
2
x
=
3
−
l
o
g
2
x
are :
A
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is
A
1
√
5
−
log
2
x
=
3
−
log
2
x
---- ( 1 )
Let
log
2
x
=
y
----- ( 2 )
⇒
√
5
−
y
=
3
−
y
Squaring both sides,
⇒
5
−
y
=
9
−
6
y
+
y
2
⇒
y
2
−
5
y
+
4
=
0
⇒
y
2
−
4
y
−
y
+
4
=
0
⇒
y
(
y
−
4
)
−
1
(
y
−
4
)
=
0
⇒
(
y
−
4
)
(
y
−
1
)
=
0
⇒
y
=
4
and
y
=
1
Substituting
y
=
1
in ( 1 ) we get,
⇒
log
2
x
=
1
We know,
log
b
a
=
x
⇒
a
=
b
x
∴
x
=
2
1
=
2
Substituting
y
=
4
in ( 1 ) we get,
⇒
log
2
x
=
4
We know,
log
b
a
=
x
⇒
a
=
b
x
∴
x
=
2
4
=
16
Substituting
log
2
x
=
1
in ( 1 ) we get,
√
5
−
1
=
3
−
1
⇒
√
4
=
2
∴
2
=
2
Substituting
log
2
x
=
4
in ( 1 ) we get,
√
5
−
4
=
3
−
4
⇒
√
1
=
−
1
∴
1
=
−
1
Hence, we can see only
log
2
x
=
1
satisfying.
∴
√
5
−
log
2
x
=
3
−
log
2
x
has only
1
solution.
Suggest Corrections
0
Similar questions
Q.
Number of solution(s) satisfying the equation,
3
x
2
−
2
x
3
=
log
2
(
x
2
+
1
)
−
log
2
x
is:
Q.
The number of solution(s) of
log
4
(
x
−
1
)
=
log
2
(
x
−
3
)
is/are :
Q.
The number of solutions of
log
2
(
x
+
5
)
=
6
−
x
is
Q.
The number of integral solutions satisfying the equation
√
log
2
x
−
1
−
1
2
log
2
(
x
3
)
+
2
>
0
is equal to
Q.
The number of solutions of
log
4
(
x
−
1
)
=
log
2
(
x
−
3
)
will be
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
Domestic Electric Circuits
MATHEMATICS
Watch in App
Explore more
Properties of Iota
Standard XII Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
AI Tutor
Textbooks
Question Papers
Install app