Aptitude - Logarithm - Discussion
Discussion Forum : Logarithm - General Questions (Q.No. 5)
5.
| If log | a | + | log | b | = log (a + b), then: |
| b | a |
Answer: Option
Explanation:
| log | a | + log | b | = log (a + b) |
| b | a |
 log (a + b) = log |
 |
a | x | b |  |
= log 1. |
| b | a |
So, a + b = 1.
Discussion:
33 comments Page 1 of 4.
Charu said:
1 decade ago
I can't got it. Please explain it more clearly.
Divya said:
1 decade ago
I can't understand your point. Can any one please explain me?
Radhe.. said:
1 decade ago
'+' indicate the multiplication in log.and '-' indicate the division in log....
Siddhu said:
1 decade ago
We know that: log a + log b = log (a+b);
Similarly log(a/b) + log(b/a) = log (a/b * b/a);
We get log 1 bcoz all terms will get cancels
and log 1 = 0;
This is about L.H.S. So look at the R.H.S log(a+b) is there and we have to get R.H.S also zero it is possible when we get log 1 = log(a+b) so a+b = 1.
Similarly log(a/b) + log(b/a) = log (a/b * b/a);
We get log 1 bcoz all terms will get cancels
and log 1 = 0;
This is about L.H.S. So look at the R.H.S log(a+b) is there and we have to get R.H.S also zero it is possible when we get log 1 = log(a+b) so a+b = 1.
Priya said:
1 decade ago
Thanks siddhu.
Raghavendra said:
1 decade ago
log(a+b)=log(a/b)+log(b/a)
log(a+b)=l0ga-logb+logb-loga
log(a+b)=0
a+b=10^0
so
a+b=1
log(a+b)=l0ga-logb+logb-loga
log(a+b)=0
a+b=10^0
so
a+b=1
MADHU said:
1 decade ago
Can any one explain how log(a+b)=loga+logb
Sameer said:
1 decade ago
Here someone wrote: loga+logb = log(a+b).
But how log a/b + log b/a = log (a/b x b/a)?
Why not like this log(a/b + b/a)?
But how log a/b + log b/a = log (a/b x b/a)?
Why not like this log(a/b + b/a)?
Keshav said:
1 decade ago
@Sameer according to the logarithmic identity.
log A + log B = log(A*B).
& log A + log B = log(A+B).
There is no such identity.
Please check.
log A + log B = log(A*B).
& log A + log B = log(A+B).
There is no such identity.
Please check.
Manikantan said:
1 decade ago
log a/b +log b/a = log a - log b + log b - log a = 0.
log (a+b)=0 => a+b =1 *(log 1 =0).
log (a+b)=0 => a+b =1 *(log 1 =0).
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