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 3 of 4.
4lick said:
8 years ago
loga + logb = loga/b.
Preetiranjan panda said:
8 years ago
@Shree.
log{(a+b)/3}=0.5(log a +log b).
=1/2(log a+log b),
=1/2 log a + 1/2 log b,
= log a^1/2 + log b^1/2,
=log (a^1/2 * b^1/2),
=>(a+b)/3 = a^1/2 * b^1/2,
=(ab)^1/2.
=>{(a+b)/3}^2 = ab,
=>(a+b)^2 /3^2= ab,
=>(a+b)^2 /9 = ab,
=>(a+b)^2 = 9ab,
=> a^2 +b^2+2ab = 9ab,
=> a^2+b^2 = 9ab - 2ab,
=> a^2+b^2 = 7ab.
(Proved)
log{(a+b)/3}=0.5(log a +log b).
=1/2(log a+log b),
=1/2 log a + 1/2 log b,
= log a^1/2 + log b^1/2,
=log (a^1/2 * b^1/2),
=>(a+b)/3 = a^1/2 * b^1/2,
=(ab)^1/2.
=>{(a+b)/3}^2 = ab,
=>(a+b)^2 /3^2= ab,
=>(a+b)^2 /9 = ab,
=>(a+b)^2 = 9ab,
=> a^2 +b^2+2ab = 9ab,
=> a^2+b^2 = 9ab - 2ab,
=> a^2+b^2 = 7ab.
(Proved)
Dimpal said:
8 years ago
Not getting this point, please explain clearly.
Nams said:
8 years ago
a/b+b/a=ab.
a^2-2ab+b^2=0,
(a-b)^2=
a-b=0,
a=b.
a^2-2ab+b^2=0,
(a-b)^2=
a-b=0,
a=b.
Shubham said:
7 years ago
Thanks all for giving the solution.
Kevin augustine said:
7 years ago
Log a/b + log b/a = log(a+b).
Then,
log a - log b + log b - log a = log (a + b).
Lhs cancelled each other so lhs = 0.
Then,
log (a + b) = 0.
So a+b should be 1.
Answer is 1.
Then,
log a - log b + log b - log a = log (a + b).
Lhs cancelled each other so lhs = 0.
Then,
log (a + b) = 0.
So a+b should be 1.
Answer is 1.
(1)
Patil said:
6 years ago
I cannot understand please explain the method in detail.
Pradeep Sharma said:
6 years ago
Here we have;
Log(a/b) + Log (b/a) = Log (a + b ) --> (1)
So, by taking LHS.
we have,
Log (a/b * b/a).
So, now we have Log 1 (as a/b * b/a =1).
From equation 1 we have;
Log 1 = Log (a+b).
So by this we have,
a+b =1.
Hence proved.
Log(a/b) + Log (b/a) = Log (a + b ) --> (1)
So, by taking LHS.
we have,
Log (a/b * b/a).
So, now we have Log 1 (as a/b * b/a =1).
From equation 1 we have;
Log 1 = Log (a+b).
So by this we have,
a+b =1.
Hence proved.
(2)
Madhav said:
6 years ago
log(a/b) + log(b/a) = log(a+b).
>loga - logb + logb-loga = log(a+b),
>0 = log(a+b),
>log1 = log(a+b).
Therefore from LHS AND RHS.
a+b=1 ***log1 = 0.
>loga - logb + logb-loga = log(a+b),
>0 = log(a+b),
>log1 = log(a+b).
Therefore from LHS AND RHS.
a+b=1 ***log1 = 0.
(9)
Md.anas said:
6 years ago
log(a/b)+log(b/a) = log(a+b)
log(a-b)+log(b-a) = log(a+b)
log(a-b)+log-(a-b) = log(a+b)
log(a-b)+log(1/a-b) = log(a+b)
log(a-b/a-b) = log(a+b)
log(1) = log(a+b)
1 = a+b
a+b = 1.
log(a-b)+log(b-a) = log(a+b)
log(a-b)+log-(a-b) = log(a+b)
log(a-b)+log(1/a-b) = log(a+b)
log(a-b/a-b) = log(a+b)
log(1) = log(a+b)
1 = a+b
a+b = 1.
(9)
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