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if (log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) then abc = ? a)0 b)1 c)2 d)3
1(log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b)=tlog 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-tblog abc = 0log abc = log 1 abc = 1
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given that loga/b-c=logb/c-a=logc/a-bneed to find:abc=?so,let us take some constant, say it as x, we know that,loga/b-c=logb/c-a=logc/a-bloga=x(b-c)--------(1)logb=x(c-a)--------(2)logc=x(a-b)---------(3)add (1)+(2)+(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=0logabc=log1 (log1=0)abc=1
Hello StudentGiven(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) = kloga = k(b-c)..........(2)logb = k(c-a)..........(3)logc = k(a-b)..........(4)add eq. (1)+(2)+(3), we get,loga + logb + logc = k(b-c) + k(c-a) + k(a-b)logabc = kb – kc + kc - ka + ka - kblogabc = 0logabc = log1abc = 1Hence option B is correct.I hope this answer will help you.
Dear student,Please find the attached solution to your problem. 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) add eq. (1)+(2)+(3), we get,loga + logb + logc = k(b-c) + k(c-a) + k(a-b)or, log(abc) = kb – kc + kc - ka + ka - kbor, log(abc) = 0or, log(abc) = log1Hence, abc = 1Hence option B is correct. Hope it helps.Thanks and regards,Kushagra
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