Grade 9Algebra
if (log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) then abc = ?
a)0
b)1
c)2
d)3
if (log a) / (b-c) = (log b) / (c-a) = (log c) / (a-b) then abc = ?
a)0
b)1
c)2
d)3
a)0
b)1
c)2
d)3
6 Answers
Shanmukha
- 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
grenade
1
Rohan Gajjar
1
meghana
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)
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=0
logabc=log1 (log1=0)
abc=1
Yash Chourasiya
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)
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 - kb
logabc = 0
logabc = log1
abc = 1
Hence option B is correct.
I hope this answer will help you.
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)
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 - kb
logabc = 0
logabc = log1
abc = 1
Hence option B is correct.
I hope this answer will help you.
Kushagra Madhukar
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 - kb
or, log(abc) = 0
or, log(abc) = log1
Hence, abc = 1
Hence option B is correct.
Hope it helps.
Thanks and regards,
Kushagra
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