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4-1-x-6-1-x-9-1-x-x-R-

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Question Number 187672 by Humble last updated on 20/Feb/23
4^(−(1/x)) +6^(−(1/x)) = 9^(−(1/x)) x∈R
$$\mathrm{4}^{−\frac{\mathrm{1}}{{x}}} +\mathrm{6}^{−\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{9}^{−\frac{\mathrm{1}}{{x}}} \\ $$$${x}\in\mathbb{R} \\ $$
Answered by SEKRET last updated on 20/Feb/23
(4^(− (1/x)) /9^(− (1/x)) ) + (6^((−1)/x) /9^((−1)/x) ) =1 ((4/9))^((−1)/x) +((6/9))^((−1)/x) =1 ((2/3))^(2∙ ((−1)/x)) + ((2/3))^((−1)/x) =1 t^2 +t=1 t^2 +2∙(1/2)∙t+(1/4)=1+(1/4) (t+(1/2))^2 = (5/4) t= ((−1∓(√5))/2) ((2/3))^((−1)/x) = (((√(5 )) −1)/2) ((−1)/x)= log_(((2/3))) ((((√5) −1)/2)) (1/x) =log_(((3/2))) ((((√5) −1)/2)) x= log_(((((√5) −1)/2))) ((3/2))
$$\:\:\frac{\mathrm{4}^{−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} }{\mathrm{9}^{−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} }\:+\:\frac{\mathrm{6}^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} }{\mathrm{9}^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} }\:=\mathrm{1} \\ $$$$\:\:\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} +\left(\frac{\mathrm{6}}{\mathrm{9}}\right)^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} =\mathrm{1} \\ $$$$\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}\centerdot\:\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} +\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} =\mathrm{1} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}=\mathrm{1} \\ $$$$\:\:\:\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\left(\boldsymbol{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\boldsymbol{\mathrm{t}}=\:\:\frac{−\mathrm{1}\mp\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}} =\:\frac{\sqrt{\mathrm{5}\:}\:\:−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{−\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\:\boldsymbol{\mathrm{log}}_{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} \left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:=\boldsymbol{\mathrm{log}}_{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{x}}=\:\boldsymbol{\mathrm{log}}_{\left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}\right)} \left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$
Commented by Humble last updated on 20/Feb/23
nice solution, sir
$${nice}\:{solution},\:\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 20/Feb/23
4^(−(1/x)) +6^(−(1/x)) = 9^(−(1/x)) ((4/6))^(−1/x) +1=((9/6))^(−1/x) ((3/2))^(1/x) +1=((2/3))^(1/x) y+1=(1/y) ;y≥0 y^2 +y−1=0 y=((−1+(√(1+4)))/2) log((3/2))^(1/x) =log (((−1+(√5))/2)) ((1/x))log((3/2))=log(−1+(√5) )−log2 x=((log3−log2 )/(log(−1+(√5) )−log2))
$$\mathrm{4}^{−\frac{\mathrm{1}}{{x}}} +\mathrm{6}^{−\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{9}^{−\frac{\mathrm{1}}{{x}}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{−\mathrm{1}/{x}} +\mathrm{1}=\left(\frac{\mathrm{9}}{\mathrm{6}}\right)^{−\mathrm{1}/{x}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} +\mathrm{1}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}/{x}} \\ $$$${y}+\mathrm{1}=\frac{\mathrm{1}}{{y}}\:\:;{y}\geqslant\mathrm{0} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} =\mathrm{log}\:\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{5}}\:\right)−\mathrm{log2} \\ $$$${x}=\frac{\mathrm{log3}−\mathrm{log2}\:\:}{\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{5}}\:\right)−\mathrm{log2}}\: \\ $$$$ \\ $$
Commented by Humble last updated on 20/Feb/23
Excellent!!
$${Excellent}!! \\ $$

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