Answers
Answered by
0
Answer:
Let
log
a
b
−
c
=
log
b
c
−
a
=
log
c
a
−
b
=
k
then considering 10 base logarithm we get
a
=
10
k
(
b
−
c
)
b
=
10
k
(
c
−
a
)
c
=
10
k
(
a
−
b
)
So
a
a
=
10
k
(
a
b
−
c
a
)
b
b
=
10
k
(
b
c
−
a
b
)
c
c
=
10
k
(
c
a
−
b
c
)
Hence the numerical value of
a
a
⋅
b
b
⋅
c
c
=
10
k
(
a
b
−
c
a
)
⋅
10
k
(
b
c
−
a
b
)
⋅
10
k
(
c
a
−
b
c
)
=
10
k
(
a
b
−
c
a
+
b
c
−
a
b
+
c
a
−
b
c
)
=
10
0
=
1
Similar questions