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Question Number 172084 by Mikenice last updated on 23/Jun/22

$$solve$$$${\left(\sqrt{5+\sqrt{24}}\right)}^x-10={\left(\sqrt{5-\sqrt{24}}\right)}^x$$

Answered by mr W last updated on 23/Jun/22

$$lett={\left(\sqrt{5+\sqrt{24}}\right)}^x>0$$$${\left(\sqrt{5-\sqrt{24}}\right)}^x={\left(\frac{1}{\sqrt{5+\sqrt{24}}}\right)}^x=\frac{1}{t}$$$$t-10=\frac{1}{t}$$$$t^2-10t-1=0$$$$t=5+\sqrt{26}(5-\sqrt{26}<0rejected)$$$${\left(\sqrt{5+\sqrt{24}}\right)}^x=5+\sqrt{26}$$$$\frac{x}{2}\times \ln (5+\sqrt{24})=\ln (5+\sqrt{26})$$$$\Rightarrow x=\frac{2\ln (5+\sqrt{26})}{\ln (5+\sqrt{24})}$$

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