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Question Number 191103 by Mingma last updated on 18/Apr/23

Answered by Rasheed.Sindhi last updated on 18/Apr/23

$$4^{-\frac{1}{x}}+6^{-\frac{1}{x}}=9^{-\frac{1}{x}}$$$$\frac{2^{-2/x}}{2^{-1/x}\cdot 3^{-1/x}}+\frac{2^{-1/x}\cdot 3^{-1/x}}{2^{-1/x}\cdot 3^{-1/x}}=\frac{3^{-2/x}}{2^{-1/x}\cdot 3^{-1/x}}$$$$\frac{2^{-2/x+1/x}}{3^{-1/x}}+1=\frac{3^{-2/x+1/x}}{2^{-1/x}}$$$$\frac{3^{1/x}}{2^{1/x}}+1=\frac{2^{1/x}}{3^{1/x}}$$$$y+1=\frac{1}{y}$$$$y^2+y-1=0$$$$y=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$$$$\frac{3^{1/x}}{2^{1/x}}=\frac{-1\pm \sqrt{5}}{2}$$$${\left(\frac{3}{2}\right)}^{1/x}=\frac{-1\pm \sqrt{5}}{2}$$$$\left(\frac{1}{x}\right)\log \left(\frac{3}{2}\right)=\log \left(\frac{-1+\sqrt{5}}{2}\right)$$$$[\because \log \left(\frac{-1-\sqrt{5}}{2}\right)\notin \mathbb{R}]$$$$\frac{1}{x}=\frac{\log (-1\times \sqrt{5})-\log 2}{\log 3-\log 2}$$$$x=\frac{\log 3-\log 2}{\log (-1+\sqrt{5})-\log 2}$$$$x=\frac{{\log }_{2}3-1}{{\log }_2(1+\sqrt{5})-1}$$

Commented by Mingma last updated on 18/Apr/23

Nice solution, sir!

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