Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 167550 by mathlove last updated on 19/Mar/22

Answered by MJS_new last updated on 19/Mar/22

$$\mathrm{pythagorean}\mathrm{triple}{5}^2+{12}^2={13}^2$$$$\Rightarrow x=4$$

Answered by Rasheed.Sindhi last updated on 19/Mar/22

$$5^{\sqrt{x}}+{12}^{\sqrt{x}}={13}^{\sqrt{x}}$$$${\left(\frac{5}{13}\right)}^{\sqrt{x}}+{\left(\frac{12}{13}\right)}^{\sqrt{x}}=1$$$$Let\overline{)\begin{array}{c}\frac{5}{13}=\sin \theta \end{array}}$$$${\sin }^2\theta =\frac{25}{169}\Rightarrow 1-{\cos }^2\theta =\frac{25}{169}$$$$\Rightarrow {\cos }^2\theta =1-\frac{25}{169}=\frac{144}{169}\Rightarrow \overline{)\begin{array}{c}\frac{12}{13}=\cos \theta \end{array}}$$$${\left(\sin \theta \right)}^{\sqrt{x}}+{\left(\cos \theta \right)}^{\sqrt{x}}=1$$$${\left(\sin \theta \right)}^{\sqrt{x}}+{\left(\cos \theta \right)}^{\sqrt{x}}={\sin }^2\theta +{\cos }^2\theta$$$$Comparingexponentsonbothsides$$$$\sqrt{x}=2$$$$x=4$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com