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8-x-x-x-2-Show-that-x-4-




Question Number 7655 by Tawakalitu. last updated on 07/Sep/16
((8/x))^x  =  x^2     Show that  x = 4
$$\left(\frac{\mathrm{8}}{{x}}\right)^{{x}} \:=\:\:{x}^{\mathrm{2}} \\ $$$$ \\ $$$${Show}\:{that}\:\:{x}\:=\:\mathrm{4} \\ $$
Commented by FilupSmith last updated on 07/Sep/16
trial and error  x=3  ((8/3))^3 =3^2   (2.6^− )^3 ≠9      (2.6^− =2+0.6^−  →2^3 =8,                   n=0.6^− , for n>1,   (1/n^k )<0⇒∴2.6^− ≠9)  ((8/4))^4 =4^2   2^4 =4^2
$$\mathrm{trial}\:\mathrm{and}\:\mathrm{error} \\ $$$${x}=\mathrm{3} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}.\overset{−} {\mathrm{6}}\right)^{\mathrm{3}} \neq\mathrm{9}\:\:\:\:\:\:\left(\mathrm{2}.\overset{−} {\mathrm{6}}=\mathrm{2}+\mathrm{0}.\overset{−} {\mathrm{6}}\:\rightarrow\mathrm{2}^{\mathrm{3}} =\mathrm{8},\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{n}=\mathrm{0}.\overset{−} {\mathrm{6}},\:{for}\:{n}>\mathrm{1},\:\:\:\frac{\mathrm{1}}{{n}^{{k}} }<\mathrm{0}\Rightarrow\therefore\mathrm{2}.\overset{−} {\mathrm{6}}\neq\mathrm{9}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{4}}\right)^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} \\ $$
Answered by FilupSmith last updated on 08/Sep/16
((8/x))^x =x^2   2^x =x^2   (8/x)=2  8=2x  x=4
$$\left(\frac{\mathrm{8}}{{x}}\right)^{{x}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}} \\ $$$$\frac{\mathrm{8}}{{x}}=\mathrm{2} \\ $$$$\mathrm{8}=\mathrm{2}{x} \\ $$$${x}=\mathrm{4} \\ $$
Commented by Tawakalitu. last updated on 09/Sep/16
Thanks
$${Thanks} \\ $$
Answered by FilupSmith last updated on 08/Sep/16
((8/x))^x  =  x^2   xln((8/x))=2ln(x)  ln(8)−ln(x)=(2/x)ln(x)  ln(8)=((2/x)+1)ln(x)  ln(2^3 )=((2/x)+1)ln(x)  ln(2^(2((3/2))) )=((2/x)+1)ln(x)  (3/2)ln(2^2 )=((2/x)+1)ln(x)  (3/2)ln(4)=((2/x)+1)ln(x)  x=4  ⇒  (2/4)+1=(3/2)  LHS=RHS
$$\left(\frac{\mathrm{8}}{{x}}\right)^{{x}} \:=\:\:{x}^{\mathrm{2}} \\ $$$${x}\mathrm{ln}\left(\frac{\mathrm{8}}{{x}}\right)=\mathrm{2ln}\left({x}\right) \\ $$$$\mathrm{ln}\left(\mathrm{8}\right)−\mathrm{ln}\left({x}\right)=\frac{\mathrm{2}}{{x}}\mathrm{ln}\left({x}\right) \\ $$$$\mathrm{ln}\left(\mathrm{8}\right)=\left(\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$$\mathrm{ln}\left(\mathrm{2}^{\mathrm{3}} \right)=\left(\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$$\mathrm{ln}\left(\mathrm{2}^{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \right)=\left(\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}^{\mathrm{2}} \right)=\left(\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{4}\right)=\left(\frac{\mathrm{2}}{{x}}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$${x}=\mathrm{4}\:\:\Rightarrow\:\:\frac{\mathrm{2}}{\mathrm{4}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${LHS}={RHS} \\ $$
Commented by Tawakalitu. last updated on 09/Sep/16
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$

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