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 2 of 4.
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.
Dimpal said:
8 years ago
Not getting this point, please explain clearly.
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)
4lick said:
8 years ago
loga + logb = loga/b.
Harsh said:
8 years ago
@Siddharth.
You cancelled out (a-b) from both sides, which is not accepted since a-b=0 and division by 0 is undefined?
You cancelled out (a-b) from both sides, which is not accepted since a-b=0 and division by 0 is undefined?
Biswas said:
9 years ago
The, log(2+3) = log(2 * 3) should be same in the question 1(b).
Vijaylaxmi said:
9 years ago
According to rules of logarithms.
Log (a+b) =loga*logb.
Log (a/b) =loga-logb.
Log (a+b) =loga*logb.
Log (a/b) =loga-logb.
Charu said:
1 decade ago
I can't got it. Please explain it more clearly.
Amith said:
9 years ago
log (a + b + c) = log a + log b + log c is true only when a + b + c = a * b * c
Eg: 1, 2, 3.
Eg: 1, 2, 3.
Ravi said:
10 years ago
Formula for log x to base a.
Can anyone help me?
Can anyone help me?
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