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let us consider every ratio is equal to some constant k
loga/b-c=logb/c-a=logc/a-b=k
then
loga=k(b-c)
logb=k(c-a)
logc=k(a-b)
now multiply equation by a,b,c respestively
aloga=k(ab-ac)
blogb=k(bc-ab)
clogc=k(ac-bc)
add all the three equations
we get
aloga+blogb+clogc=k(ab-ac+bc-ab+ac-bc)
log(aabbcc)=0
s0
aabbcc =1
simple right
plzzz approve it means a lot to me
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