Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

$${Thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com