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

$$Given\log {}_3\left(n\right)=\log {}_6\left(m\right)=\log {}_{12}(m+n)$$$$\frac{m}{n}=?$$

Answered by bobhans last updated on 21/Nov/21

$$\log {}_3\left(\mathrm{n}\right)=\mathrm{k}\Rightarrow \mathrm{n}=3^{\mathrm{k}}$$$$\log {}_6\left(\mathrm{m}\right)=\mathrm{k}\Rightarrow \mathrm{m}=6^{\mathrm{k}}$$$$\log {}_{12}(\mathrm{m}+\mathrm{n})=\mathrm{k}\Rightarrow \mathrm{m}+\mathrm{n}={12}^{\mathrm{k}}$$$$\Rightarrow 3^{\mathrm{k}}+6^{\mathrm{k}}={12}^{\mathrm{k}};\frac{\mathrm{m}}{\mathrm{n}}=\frac{6^{\mathrm{k}}}{3^{\mathrm{k}}}=2^{\mathrm{k}}$$$$\Rightarrow {\left(\frac{1}{4}\right)}^{\mathrm{k}}+{\left(\frac{1}{2}\right)}^{\mathrm{k}}=1$$$$\Rightarrow \left(2^{-2\mathrm{k}}\right)+\left(2^{-\mathrm{k}}\right)=1;{\left(2^{-\mathrm{k}}\right)}^2+\left(2^{-\mathrm{k}}\right)-1=0$$$$\Rightarrow 2^{-\mathrm{k}}=\frac{-1+\sqrt{5}}{2}\Rightarrow 2^{\mathrm{k}}=\frac{2}{\sqrt{5}-1}$$$$\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2(\sqrt{5}+1)}{4}=\frac{\sqrt{5}+1}{2}$$$$$$

Commented by Tony6400 last updated on 21/Nov/21

Commented by cortano last updated on 21/Nov/21

$$2^k>0forx\in \mathbb{R}$$

Commented by cortano last updated on 21/Nov/21

$$since\log {}_3\left(n\right)\Rightarrow definedonn>0$$$$so{3}^k>0$$

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