Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Let A =(3√ log 3 2)√ log 2 3+(3 log 3 2)√ log 2 3+(3√ log 3 2)√ log 3 2 and B=(2√ log 3 2)√ log 2 3+(2 log 2 3)√ log 3 2-(2√ log 2 3)√ log 2 3. Then find the value of (A-B).
Q. Let
A
=
(
3
l
o
g
3
2
)
l
o
g
2
3
+
(
3
l
o
g
3
2
)
l
o
g
2
3
+
(
3
l
o
g
3
2
)
l
o
g
3
2
and
B
=
(
2
l
o
g
3
2
)
l
o
g
2
3
+
(
2
l
o
g
2
3
)
l
o
g
3
2
−
(
2
l
o
g
2
3
)
l
o
g
2
3
. Then find the value of
(
A
−
B
)
.
37
100
Continuity and Differentiability
Report Error
Answer:
6
Solution:
A
=
3
1
+
2
l
o
g
2
3
+
2
;
B
=
2
+
3
l
o
g
3
2
−
3
Hence,
(
A
−
B
)
=
6