Aptitude - Logarithm - Discussion

Discussion :: Logarithm - General Questions (Q.No.5)

5. 

If log a + log b = log (a + b), then:
b a

[A]. a + b = 1
[B]. a - b = 1
[C]. a = b
[D]. a2 - b2 = 1

Answer: Option A

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.


Charu said: (Nov 19, 2010)  
I can't got it. Please explain it more clearly.

Divya said: (Jan 10, 2011)  
I can't understand your point. Can any one please explain me?

Radhe.. said: (May 15, 2011)  
'+' indicate the multiplication in log.and '-' indicate the division in log....

Siddhu said: (Aug 10, 2011)  
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.

Priya said: (Sep 1, 2011)  
Thanks siddhu.

Raghavendra said: (Dec 23, 2011)  
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

Madhu said: (Sep 1, 2012)  
Can any one explain how log(a+b)=loga+logb

Sameer said: (Jul 6, 2013)  
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)?

Keshav said: (Jul 25, 2013)  
@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.

Manikantan said: (Dec 19, 2013)  
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).

Rajaa said: (Apr 19, 2014)  
Problem: log(a/b)+log(b/a) = log(a+b).

log(a)+log(b) = log(a*b).

So LHS log(a/b)+log(b/a) = log(a/b*b/a).
= log(1).

And RHS log(a+b).

LHS = RHS.
log(a+b) = log(1).
Hence a+b = 1.

Dilshan Fernando said: (Mar 11, 2015)  
log(a/b)+log(b/a) = log(a+b).

log(a/b*b/a) = log(a+b).

log 1 = log (a+b).

log 1 = 0.

So that a+b = 0.

a = -b.

Shree said: (Jul 11, 2015)  
What is the solution for:

log{(a+b)/3} = 0.5(log a + log b).

The answer is a^2+b^2 = 7ab.

But I want the procedure using the formula as mentioned here please.

Shree said: (Jul 11, 2015)  
log x = (1/2) log y = (1/5) log z, the value of x4y3z-2 is?

Can anyone clearly explain step by step please?

Ravi said: (Dec 9, 2015)  
Formula for log x to base a.

Can anyone help me?

Amith said: (Sep 10, 2016)  
log (a + b + c) = log a + log b + log c is true only when a + b + c = a * b * c

Eg: 1, 2, 3.

Siddharth said: (Jan 31, 2017)  
I have a question.

Log a =log b
a = b
a * a = ab
(a * a)- b * b = ab - b *b
(a + b)(a - b) = b(a - b)
a + b = b
But a = b
So
B + b = b
Were is the mistake? Please clarify this.

Vijaylaxmi said: (Feb 5, 2017)  
According to rules of logarithms.

Log (a+b) =loga*logb.
Log (a/b) =loga-logb.

Biswas said: (Feb 8, 2017)  
The, log(2+3) = log(2 * 3) should be same in the question 1(b).

Harsh said: (May 26, 2017)  
@Siddharth.

You cancelled out (a-b) from both sides, which is not accepted since a-b=0 and division by 0 is undefined?

4Lick said: (Jun 14, 2017)  
loga + logb = loga/b.

Preetiranjan Panda said: (Jun 22, 2017)  
@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)

Dimpal said: (Jul 22, 2017)  
Not getting this point, please explain clearly.

Nams said: (Dec 27, 2017)  
a/b+b/a=ab.
a^2-2ab+b^2=0,
(a-b)^2=
a-b=0,
a=b.

Shubham said: (Oct 2, 2018)  
Thanks all for giving the solution.

Kevin Augustine said: (Feb 5, 2019)  
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.

Patil said: (Apr 19, 2019)  
I cannot understand please explain the method in detail.

Pradeep Sharma said: (May 21, 2019)  
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.

Madhav said: (Sep 27, 2019)  
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.

Md.Anas said: (Oct 28, 2019)  
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.

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