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Question

Question: If \[\log \left( {m + n} \right) = \log m + \log n\], then : (A) \(mn = 1\) (B) \(m = - n\) ...

If log(m+n)=logm+logn\log \left( {m + n} \right) = \log m + \log n, then :
(A) mn=1mn = 1
(B) m=nm = - n
(C) mm1=n\dfrac{m}{{m - 1}} = n
(D) mn=1\dfrac{m}{n} = 1

Explanation

Solution

Compare the log(m+n)=logm+logn\log \left( {m + n} \right) = \log m + \log n with the fundamental law of logarithm, i.e., log(mn)=logm+logn\log \left( {mn} \right) = \log m + \log n and then find a relation between mm and nn.

Complete step-by-step answer:
Given, log(m+n)=logm+logn\log \left( {m + n} \right) = \log m + \log n..........….. (1)
We know that the basic product law of logarithm is given by,
log(mn)=logm+logn\log \left( {mn} \right) = \log m + \log n...........….. (2)
On comparing (1) and (2), we get-
log(m+n)=log(mn)\log \left( {m + n} \right) = \log \left( {mn} \right)
Now, log\log on both sides cancel out and thus we get-
m+n=mnm + n = mn
m=mnn\Rightarrow m = mn - n
m=n(m1)\Rightarrow m = n\left( {m - 1} \right)
mm1=n\Rightarrow \dfrac{m}{{m - 1}} = n

Hence, option (C) is the correct answer.

Note: A logarithm can have any positive value as its base, but two log bases are more useful than the others: base-1010 and base -ee. If a log\log has no base written, we should generally assume that the base is 1010 as in our question. Also, the log\log of both sides can be cancelled, only when the bases of both the log\log are equal.