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 126619 by O Predador last updated on 22/Dec/20

$$$$$$5^{\mathbf{x}}+7^{\mathbf{x}}={(9.8)}^{\mathbf{x}}$$$$$$$$\mathbf{x}=?$$$$$$

Answered by Dwaipayan Shikari last updated on 22/Dec/20

$${\left(\frac{5}{9.8}\right)}^x+{\left(\frac{7}{9.8}\right)}^x=1$$$$\Rightarrow \left({\left(\frac{5}{7}\right)}^{x}a\right)+a=1a={\left(\frac{7}{9.8}\right)}^x={\left(\frac{10}{14}\right)}^x={\left(\frac{5}{7}\right)}^x$$$$\Rightarrow a^2+a-1=0\Rightarrow a=\frac{-1\pm \sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}$$$${\left(\frac{5}{7}\right)}^x=\frac{\sqrt{5}-1}{2}\Rightarrow x=\frac{log\left(\frac{\sqrt{5}-1}{2}\right)}{log\left(\frac{5}{7}\right)}\sim 1.4301679980845866709...$$

Commented by O Predador last updated on 22/Dec/20

$$Wonderful!$$

Answered by Olaf last updated on 22/Dec/20

$$5^x+7^x=9.8^x={\left(\frac{7\times 14}{5\times 2}\right)}^x$$$$5^x+7^x=9.8^x=7^x{\left(\frac{7}{5}\right)}^x$$$${\left(\frac{5}{7}\right)}^x+1=9.8^x={\left(\frac{7}{5}\right)}^x$$$$\mathrm{Let}\mathrm{X}={\left(\frac{5}{7}\right)}^x$$$$\mathrm{X}+1=\frac{1}{\mathrm{X}}$$$${\mathrm{X}}^2+\mathrm{X}-1=0$$$$\mathrm{X}=\frac{-1\pm \sqrt{5}}{2}$$$$\frac{-1-\sqrt{5}}{2}\mathrm{impossible}\mathrm{because}\mathrm{X}>0$$$$\mathrm{X}=\frac{-1+\sqrt{5}}{2}={\left(\frac{5}{7}\right)}^x$$$$x\ln \frac{5}{7}=\ln \left(\frac{\sqrt{5}-1}{2}\right)$$$$x=\frac{\ln \left(\frac{\sqrt{5}-1}{2}\right)}{\ln \frac{5}{7}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com