do the question 13.picure enclosed
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we know logarithm concept
log (a base b)=1/log (b base a)
use here this concept
1/(log5 x)(log4 x)+1/(log5 x)(log3 x)=1/(log3 x )(log4 x )(log5 x) -1/(log3 x )(log4 x)
let
log3 x=p
log4 x =q
log5 x=r
hence
1/qr+1/rp=1/pqr - 1/pq
1/qr +1/pq+1/pr=1/pqr
(p+q+r)/pqr= 1/pqr
p+q+r=1
hence
log3 x + log4 x + log 5 x=1
log60 x=1
x=60
log (a base b)=1/log (b base a)
use here this concept
1/(log5 x)(log4 x)+1/(log5 x)(log3 x)=1/(log3 x )(log4 x )(log5 x) -1/(log3 x )(log4 x)
let
log3 x=p
log4 x =q
log5 x=r
hence
1/qr+1/rp=1/pqr - 1/pq
1/qr +1/pq+1/pr=1/pqr
(p+q+r)/pqr= 1/pqr
p+q+r=1
hence
log3 x + log4 x + log 5 x=1
log60 x=1
x=60
abhi178:
I hope this is right answer
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