Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 27718 by mondodotto@gmail.com last updated on 13/Jan/18

Answered by prakash jain last updated on 13/Jan/18

((√(2+(√3))) )^x =a ((√(2−(√3))) )^x =(1/a) a+(1/a)=4⇒a^2 −4a+1=0 a=((4±(√(12)))/2)=2±(√3) ((√(2+(√3))))^x =(2±(√3)) ((√(2+(√3))))^x =(2+(√3))⇒x=2 ((√(2+(√3))))^x =(2−(√3))⇒x=−2 x=±2

$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} ={a} \\ $$$$\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} =\frac{\mathrm{1}}{{a}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{4}\Rightarrow{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{4}\pm\sqrt{\mathrm{12}}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\Rightarrow{x}=\mathrm{2} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\Rightarrow{x}=−\mathrm{2} \\ $$$${x}=\pm\mathrm{2} \\ $$

Commented by mondodotto@gmail.com last updated on 13/Jan/18

thanx a lot sir

$$\mathrm{thanx}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com