# 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

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.

 Repith said: (Dec 2, 2020) Thank you very much for explaining this..

 Archana said: (Jan 28, 2021) Thanks @Ragavendra.