Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The value of a for which (1/ log 2 a)+(1/ log 3 a)+(1/ log 4 a)=1, is
Q. The value of a for which
l
o
g
2
a
1
+
l
o
g
3
a
1
+
l
o
g
4
a
1
=
1
, is
130
111
Continuity and Differentiability
Report Error
A
9
B
12
C
18
D
24
Solution:
l
o
g
2
a
1
+
l
o
g
3
a
1
+
l
o
g
4
a
1
=
1
⇒
lo
g
a
2
+
lo
g
a
3
+
lo
g
a
4
=
1
⇒
lo
g
a
24
=
1
⇒
a
=
24