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**Hint:**Compare the \[\log \left( {m + n} \right) = \log m + \log n\] with the fundamental law of logarithm, i.e., $\log \left( {mn} \right) = \log m + \log n$ and then find a relation between $m$ and $n$.

**Complete step-by-step answer:**

Given, \[\log \left( {m + n} \right) = \log m + \log n\]..........….. (1)

We know that the basic product law of logarithm is given by,

$\log \left( {mn} \right) = \log m + \log n$...........….. (2)

On comparing (1) and (2), we get-

$\log \left( {m + n} \right) = \log \left( {mn} \right)$

Now, $\log $on both sides cancel out and thus we get-

$m + n = mn$

$ \Rightarrow m = mn - n$

$ \Rightarrow m = n\left( {m - 1} \right)$

$ \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-$10$ and base -$e$. If a $\log $ has no base written, we should generally assume that the base is $10$ as in our question. Also, the $\log $of both sides can be cancelled, only when the bases of both the $\log $ are equal.

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