Question Number 27718 by mondodotto@gmail.com last updated on 13/Jan/18
Answered by prakash jain last updated on 13/Jan/18
$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} ={a} \\ $$$$\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} =\frac{\mathrm{1}}{{a}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{4}\Rightarrow{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{4}\pm\sqrt{\mathrm{12}}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\Rightarrow{x}=\mathrm{2} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\Rightarrow{x}=−\mathrm{2} \\ $$$${x}=\pm\mathrm{2} \\ $$
Commented by mondodotto@gmail.com last updated on 13/Jan/18
$$\mathrm{thanx}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir} \\ $$