If a, b, c are distinct positive numbers different from 1
such that then abc=
(logb a.logca-loga a) + (loga blog. b-logo b) +
(loga c.logp C-log. C) = 0
a) O
b) 1 c)-1 d)2
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Answer:
We have,
[log
b
alog
c
a−log
a
a]+[log
a
blog
c
b−log
b
b]+[log
a
clog
b
c−log
c
c]=0
[log
b
alog
c
a−1]+[log
a
blog
c
b−1]+[log
a
clog
b
c−1]=0
log
b
alog
c
a+log
a
blog
c
b+log
a
clog
b
c−3=0
log
b
alog
c
a+log
a
blog
c
b+log
a
clog
b
c=3
logb
loga
×
logc
loga
+
loga
logb
×
logc
logb
+
loga
logc
×
logb
logc
=3
logblogc
(loga)
2
+
logalogc
(logb)
2
+
logalogb
(logc)
2
=3
(loga)
3
+(logb)
3
+(logc)
3
=3(logalogblogc)
(loga)
3
+(logb)
3
+(logc)
3
−3(logalogblogc)=0
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