Question Number 191103 by Mingma last updated on 18/Apr/23
Answered by Rasheed.Sindhi last updated on 18/Apr/23
![4^(−(1/x)) +6^(−(1/x)) =9^(−(1/x)) (2^(−2/x) /(2^(−1/x) ∙3^(−1/x) ))+((2^(−1/x) ∙3^(−1/x) )/(2^(−1/x) ∙3^(−1/x) ))=(3^(−2/x) /(2^(−1/x) ∙3^(−1/x) )) (2^(−2/x + 1/x) /3^(−1/x) )+1=(3^(−2/x + 1/x) /2^(−1/x) ) (3^(1/x) /2^(1/x) )+1=(2^(1/x) /3^(1/x) ) y+1=(1/y) y^2 +y−1=0 y=((−1±(√(1+4)))/2)=((−1±(√5) )/2) (3^(1/x) /2^(1/x) )=((−1±(√5) )/2) ((3/2))^(1/x) =((−1±(√5) )/2) ((1/x))log((3/2))=log(((−1+(√5) )/2)) [∵log(((−1−(√5) )/2))∉R] (1/x)=((log(−1×(√5))−log2 )/(log 3−log 2 )) x=((log 3−log 2)/(log(−1+(√5))−log2)) x=((log_2 3−1)/(log_2 (1+(√5))−1))](Q191106.png)
$$\mathrm{4}^{−\frac{\mathrm{1}}{{x}}} +\mathrm{6}^{−\frac{\mathrm{1}}{{x}}} =\mathrm{9}^{−\frac{\mathrm{1}}{{x}}} \\ $$$$\frac{\mathrm{2}^{−\mathrm{2}/{x}} }{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}} }+\frac{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}} }{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}} }=\frac{\mathrm{3}^{−\mathrm{2}/{x}} }{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}} } \\ $$$$\frac{\mathrm{2}^{−\mathrm{2}/{x}\:+\:\mathrm{1}/{x}} }{\mathrm{3}^{−\mathrm{1}/{x}} }+\mathrm{1}=\frac{\mathrm{3}^{−\mathrm{2}/{x}\:+\:\mathrm{1}/{x}} }{\mathrm{2}^{−\mathrm{1}/{x}} } \\ $$$$\frac{\mathrm{3}^{\mathrm{1}/{x}} }{\mathrm{2}^{\mathrm{1}/{x}} }+\mathrm{1}=\frac{\mathrm{2}^{\mathrm{1}/{x}} }{\mathrm{3}^{\mathrm{1}/{x}} } \\ $$$${y}+\mathrm{1}=\frac{\mathrm{1}}{{y}} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}^{\mathrm{1}/{x}} }{\mathrm{2}^{\mathrm{1}/{x}} }=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{log}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\mathrm{log}\left(\frac{−\mathrm{1}−\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)\notin\mathbb{R}\right] \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{log}\left(−\mathrm{1}×\sqrt{\mathrm{5}}\right)−\mathrm{log2}\:\:}{\mathrm{log}\:\mathrm{3}−\mathrm{log}\:\mathrm{2}\:} \\ $$$${x}=\frac{\mathrm{log}\:\mathrm{3}−\mathrm{log}\:\mathrm{2}}{\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{log2}} \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{1}} \\ $$
Commented by Mingma last updated on 18/Apr/23
Nice solution, sir!