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

Last updated date: 18th Jun 2024
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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 $n$.

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.