# if (log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) then abc = ?a)0b)1c)2d)3

Shanmukha
35 Points
9 years ago
1. 1
(log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b)=t
log a = t(b-c)-----(1)     log b = t(c-a)-------(2)      log c = t(a-b)-------(3)
(1) + (2) + (3)
log a + log b + log c  = tb-tc+tc-ta+ta-tb
log abc = 0
log abc = log 1
abc = 1
2061 Points
9 years ago
1
Rohan Gajjar
34 Points
9 years ago
1

meghana
13 Points
7 years ago
given that
loga/b-c=logb/c-a=logc/a-b
need to find:abc=?
so,let us take some constant, say it as x,

we know that,
loga/b-c=logb/c-a=logc/a-b
loga=x(b-c)--------(1)
logb=x(c-a)--------(2)
logc=x(a-b)---------(3)
we get,
loga+logb+logc=x(b-c)+x(c-a)+x(a-b)
logabc=xb-xc+xc-xa+xa-xb                      (loga+logb+logc=logabc)
logabc=0
logabc=log1                                (log1=0)
abc=1

Yash Chourasiya
4 years ago
Hello Student

Given
(log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b)............(1)

Let eq. (1) will be equal to k
(log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) = k

loga = k(b-c)..........(2)
logb = k(c-a)..........(3)
logc = k(a-b)..........(4)

loga + logb + logc = k(b-c) + k(c-a) + k(a-b)
logabc = kb – kc + kc - ka + ka - kb
logabc = 0
logabc = log1
abc = 1

Hence option B is correct.

4 years ago
Dear student,

Given
(log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) = k (Let)
loga = k(b-c)..........(1)
logb = k(c-a)..........(2)
logc = k(a-b)..........(3)

loga + logb + logc = k(b-c) + k(c-a) + k(a-b)
or, log(abc) = kb – kc + kc - ka + ka - kb
or, log(abc) = 0
or, log(abc) = log1
Hence, abc = 1
Hence option B is correct.

Hope it helps.
Thanks and regards,
Kushagra