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

Given log _3 (n)= log _6 (m)=log _(12) (m+n) (m/n) = ?

$$\:\:\:{Given}\:\mathrm{log}\:_{\mathrm{3}} \left({n}\right)=\:\mathrm{log}\:_{\mathrm{6}} \left({m}\right)=\mathrm{log}\:_{\mathrm{12}} \left({m}+{n}\right) \\ $$$$\:\:\:\frac{{m}}{{n}}\:=\:? \\ $$

Answered by bobhans last updated on 21/Nov/21

log _3 (n)= k⇒n=3^k log _6 (m)=k⇒m=6^k log _(12) (m+n)=k⇒m+n=12^k ⇒3^k +6^k =12^k ; (m/n)=(6^k /3^k ) = 2^k ⇒ ((1/4))^k +((1/2))^k = 1 ⇒(2^(−2k) )+(2^(−k) )=1 ; (2^(−k) )^2 +(2^(−k) )−1=0 ⇒2^(−k) = ((−1+(√5))/2) ⇒2^k = (2/( (√5)−1)) ⇒ (m/n) = ((2((√5)+1))/4)= (((√5)+1)/2)

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

Commented by Tony6400 last updated on 21/Nov/21

Commented by cortano last updated on 21/Nov/21

2^k > 0 for x∈R

$$\:\mathrm{2}^{{k}} \:>\:\mathrm{0}\:{for}\:{x}\in\mathbb{R}\: \\ $$

Commented by cortano last updated on 21/Nov/21

since log _3 (n) ⇒defined on n>0 so 3^k > 0

$$\:{since}\:\mathrm{log}\:_{\mathrm{3}} \left({n}\right)\:\Rightarrow{defined}\:{on}\:{n}>\mathrm{0} \\ $$$$\:{so}\:\mathrm{3}^{{k}} \:>\:\mathrm{0} \\ $$

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