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

solve  ((√(5+(√(24)))))^x −10=((√(5−(√(24)))))^x

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

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

let t=((√(5+(√(24)))))^x >0  ((√(5−(√(24)))))^x =((1/( (√(5+(√(24)))))))^x =(1/t)  t−10=(1/t)  t^2 −10t−1=0  t=5+(√(26))    (5−(√(26)) <0 rejected)  ((√(5+(√(24)))))^x =5+(√(26))  (x/2)×ln (5+(√(24)))=ln (5+(√(26)))  ⇒x=((2 ln (5+(√(26))))/(ln (5+(√(24)))))

$${let}\:{t}=\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}}\right)^{{x}} >\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{5}−\sqrt{\mathrm{24}}}\right)^{{x}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}}}\right)^{{x}} =\frac{\mathrm{1}}{{t}} \\ $$$${t}−\mathrm{10}=\frac{\mathrm{1}}{{t}} \\ $$$${t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{5}+\sqrt{\mathrm{26}}\:\:\:\:\left(\mathrm{5}−\sqrt{\mathrm{26}}\:<\mathrm{0}\:{rejected}\right) \\ $$$$\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}}\right)^{{x}} =\mathrm{5}+\sqrt{\mathrm{26}} \\ $$$$\frac{{x}}{\mathrm{2}}×\mathrm{ln}\:\left(\mathrm{5}+\sqrt{\mathrm{24}}\right)=\mathrm{ln}\:\left(\mathrm{5}+\sqrt{\mathrm{26}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{5}+\sqrt{\mathrm{26}}\right)}{\mathrm{ln}\:\left(\mathrm{5}+\sqrt{\mathrm{24}}\right)} \\ $$

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