Given-log-3-n-log-6-m-log-12-m-n-m-n-

Question Number 159762 by cortano last updated on 21/Nov/21

G i v e n \log_{3} (n) = \log_{6} (m) = \log_{12} (m + n)

\frac{m}{n} = ?

Answered by bobhans last updated on 21/Nov/21

\log_{3} (n) = k \Rightarrow n = 3^{k}

\log_{6} (m) = k \Rightarrow m = 6^{k}

\log_{12} (m + n) = k \Rightarrow m + n = 12^{k}

\Rightarrow 3^{k} + 6^{k} = 12^{k}; \frac{m}{n} = \frac{6^{k}}{3^{k}} = 2^{k}

\Rightarrow {(\frac{1}{4})}^{k} + {(\frac{1}{2})}^{k} = 1

\Rightarrow (2^{- 2 k}) + (2^{- k}) = 1; {(2^{- k})}^{2} + (2^{- k}) - 1 = 0

\Rightarrow 2^{- k} = \frac{- 1 + \sqrt{5}}{2} \Rightarrow 2^{k} = \frac{2}{\sqrt{5} - 1}

\Rightarrow \frac{m}{n} = \frac{2 (\sqrt{5} + 1)}{4} = \frac{\sqrt{5} + 1}{2}

Commented by Tony6400 last updated on 21/Nov/21

Commented by cortano last updated on 21/Nov/21

2^{k} > 0 f o r x \in R

Commented by cortano last updated on 21/Nov/21

s i n c e \log_{3} (n) \Rightarrow d e f i n e d o n n > 0

s o 3^{k} > 0

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