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Mathematics

If log (m + n) = log m + log n, show that m=nn−1m = \dfrac{n}{n - 1}.

Logarithms

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Answer

Given,

⇒ log (m + n) = log m + log n

⇒ log (m + n) = log (mn)

⇒ (m + n) = (mn)

⇒ m - mn + n = 0

⇒ m(1 - n) + n = 0

⇒ m(1 - n) = - n

⇒ m = −n(1−n)\dfrac{-n}{(1 - n)}

⇒ m = −n−(n−1)\dfrac{-n}{-(n - 1)}

⇒ m = n(n−1)\dfrac{n}{(n - 1)}.

Hence, proved that m=nn−1m = \dfrac{n}{n - 1}.

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