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Question Number 59752 by Khairun Nisa last updated on 14/May/19

Commented by Smail last updated on 14/May/19

$${\left(\sqrt{7+\sqrt{48}}\right)}^x+{\left(\sqrt{7-\sqrt{48}}\right)}^x=14$$$${\left(\sqrt{7+\sqrt{48}}\right)}^x({\left(\sqrt{7+\sqrt{48}}\right)}^x+{\left(\sqrt{7-\sqrt{48}}\right)}^x)=14{\left(\sqrt{7+\sqrt{48}}\right)}^x$$$${\left(\sqrt{7+\sqrt{48}}\right)}^{2x}+{\left(\sqrt{(7+\sqrt{48})(7-\sqrt{48})}\right)}^x-14{\left(\sqrt{7+\sqrt{48}}\right)}^x=0$$$${\left({\left(\sqrt{7+\sqrt{48}}\right)}^x\right)}^2-14{\left(\sqrt{7+\sqrt{48}}\right)}^x+1=0$$$${({\left(\sqrt{7+\sqrt{48}}\right)}^x-7)}^2+1-49=0$$$${\left(\sqrt{7+\sqrt{48}}\right)}^x-7=\underset{-}{+}\sqrt{48}$$$${\left(\sqrt{7+\sqrt{48}}\right)}^x=7+\sqrt{48}or=7-\sqrt{48}$$$$={\left(\sqrt{7+\sqrt{48}}\right)}^{2}or={\left(\frac{1}{\sqrt{7+\sqrt{48}}}\right)}^2={\left(\sqrt{7+\sqrt{48}}\right)}^{-2}$$$$x=2or=-2$$

Commented by Khairun Nisa last updated on 14/May/19

$$ThankuSir$$

Answered by tanmay last updated on 14/May/19

$$7+\sqrt{48}$$$$7+2\sqrt{3\times 4}$$$$=2^2+{\left(\sqrt{3}\right)}^2+2\times 2\times \sqrt{3}$$$$={(2+\sqrt{3})}^2$$$${(2+\sqrt{3})}^x+{(2-\sqrt{3})}^x=14$$$$\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$$$${(2+\sqrt{3})}^x=a$$$$a+\frac{1}{a}=14$$$$a^2-14a+1=0$$$$a=\frac{14\pm \sqrt{196-4}}{2}$$$$\frac{14\pm 8\sqrt{3}}{2}=7\pm 4\sqrt{3}$$$${(2+\sqrt{3})}^x=(7+4\sqrt{3})$$$${(2+\sqrt{3})}^x={(2+\sqrt{3})}^2$$$$x=2$$$${(2+\sqrt{3})}^x=(7-4\sqrt{3})$$$${(2+\sqrt{3})}^x=\frac{1}{(7+4\sqrt{3})}$$$${(2+\sqrt{3})}^x={(2+\sqrt{3})}^{-2}$$$$x=-2$$$$sox=\pm 2$$$$$$$$$$

Commented by Khairun Nisa last updated on 14/May/19

$$ThankuSir$$

Commented by tanmay last updated on 14/May/19

$$mostwelcome$$

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