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Question Number 149008 by Samimsultani last updated on 02/Aug/21

Commented by bramlexs22 last updated on 02/Aug/21

$$\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}$$$$\sqrt{5-\sqrt{24}}=\sqrt{3}-\sqrt{2}$$$${(\sqrt{3}+\sqrt{2})}^{\mathrm{x}}-{(\sqrt{3}-\sqrt{2})}^{\mathrm{x}}=40\sqrt{6}$$$$\mathrm{let}{(\sqrt{3}+\sqrt{2})}^{\mathrm{x}}=\mathrm{u}$$

Commented by Samimsultani last updated on 02/Aug/21

$$????$$

Commented by bramlexs22 last updated on 02/Aug/21

$$\mathrm{you}\mathrm{can}\mathrm{finish}\mathrm{it}$$

Answered by Rasheed.Sindhi last updated on 02/Aug/21

$$\sqrt{5+\sqrt{24}}=t,\frac{1}{t}=\frac{1}{\sqrt{5+\sqrt{24}}}\times \frac{\sqrt{5-\sqrt{24}}}{\sqrt{5-\sqrt{24}}}$$$$=\frac{\sqrt{5-\sqrt{24}}}{\sqrt{25-24}}=\sqrt{5-\sqrt{24}}$$$$t^x-\frac{1}{t^x}=40\sqrt{6}$$$$t^{2x}-40\sqrt{6}{t}^x-1=0$$$$y^2-40\sqrt{6}y-1=0$$$$y=\frac{40\sqrt{6}\pm \sqrt{40\left(6\right)+4}}{2}$$$$y=\frac{40\sqrt{6}\pm 2\sqrt{61}}{2}=20\sqrt{6}\pm \sqrt{61}$$$$t^x=20\sqrt{6}\pm \sqrt{61}$$$$x\left(\log t\right)=\log (20\sqrt{6}\pm \sqrt{61})$$$$x=\frac{\log (20\sqrt{6}\pm \sqrt{61})}{\log \left(\sqrt{5+\sqrt{24}}\right)}$$

Answered by mr W last updated on 02/Aug/21

$$\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}$$$$\sqrt{5-\sqrt{24}}=\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}$$$$t={\left(\sqrt{5+\sqrt{24}}\right)}^x={(\sqrt{3}+\sqrt{2})}^x>0$$$$t-\frac{1}{t}=40\sqrt{6}$$$$t^2-40\sqrt{6}t-1=0$$$$t=20\sqrt{6}+\sqrt{{20}^2\times 6+1}=20\sqrt{6}+49={(\sqrt{3}+\sqrt{2})}^4$$$$x=\frac{\ln {(\sqrt{3}+\sqrt{2})}^4}{\ln (\sqrt{3}+\sqrt{2})}=4$$

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