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Question Number 176592 by Shrinava last updated on 22/Sep/22

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$2^{\mathbf{x}}+3^{\mathbf{x}}-4^{\mathbf{x}}+6^{\mathbf{x}}-9^{\mathbf{x}}=1$$

Answered by Frix last updated on 22/Sep/22

$$\mathrm{obvious}\mathrm{solution}x=0$$

Answered by a.lgnaoui last updated on 23/Sep/22

$$posonsa=2^{x}b=3^x{a}^2=2^{2x}{b}^2=3^{2x}ab=2^x{3}^x=6^x$$$$(a+b)-(a^2+b^2)+ab=1$$$$$$$$(a+b)-[{(a+b)}^2-2ab]+ab=1$$$$(a+b)-{(a+b)}^2+3ab=1\left(\mathbf{I}\right)$$$$2^x+3^x-{(2^x+3^x)}^2=1-3\times 6^x$$$$(2^x+3^x)(1-2^x-3^x)=1-3\times 6^x$$$$(2^x+3^x)(2^{2x}+3^{2x})=3\times 6^x-1$$$$(2^x+3^x)[(2^{2x}+3^{2x}+2\times 6^x)-2\times 6^x]=3\times 6^x-1$$$${(2^x+3^x)}^3-2\times 6^x(2^x+3^x)=3\times 6^x-1$$$$3ab=1+{(a+b)}^2-(a+b)$$$$3\times 6^x=1+{(2^x+3^x)}^2-(2^x+3^x)$$$${(2^x+3^x)}^3-2(2^x+3^x)\frac{1+{(2^x+3^x)}^2-(2^x+3^x)}{3}={(2^x+3^x)}^2-(2^x+3^x)$$$$onpose(2^x+3^x=z)$$$$z^3-\frac{2}{}\frac{[z(1+2z^2-z]}{3}=z^2-z$$$$z^2-\frac{2(1+2z^2-z)}{3}=z-1$$$$3z^2-4z^2+2z-2=3z-3$$$$z^2+z+1=0z=0$$$$2^x=-3^x{(-\frac{2}{3})}^x=1\Rightarrow x=0$$$$\mathrm{\Delta }=1-4={\left[\sqrt{3}i\right]}^2$$$$z1=\frac{-1\pm \sqrt{3}i}{2}$$$$2^x+3^x=\frac{1}{2}+\frac{\sqrt{3}}{2}i$$$$2(2^x+3^x)=1+\sqrt{3}i$$$$....solutionsviacoshx?$$

Commented by Shrinava last updated on 23/Sep/22

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{professor},\mathrm{yes},\mathrm{gracias}$$

Commented by Frix last updated on 23/Sep/22

$$(a+b)-(a^2+b^2)+ab=1$$$$a=u-v\wedge b=u+v$$$$2u-(2u^2+2v^2)+u^2-v^2=1$$$$2u-u^2-3v^2=1$$$$v^2=-\frac{{(u-1)}^2}{3}$$$$v=\pm \frac{u-1}{\sqrt{3}}\mathrm{i}$$$$a=u\pm \frac{\sqrt{3}(u-1)}{3}\mathrm{i}\wedge b=u\mp \frac{\sqrt{3}(u-1)}{3}\mathrm{i}$$$$2^x=u\pm \frac{\sqrt{3}(u-1)}{3}\mathrm{i}\wedge 3^x=u\mp \frac{\sqrt{3}(u-1)}{3}\mathrm{i}$$$$\mathrm{again}\mathrm{obviously}x=0\wedge u=1$$$$x=\frac{\ln (u\pm \frac{\sqrt{3}(u-1)}{3}\mathrm{i})+2n_1\pi \mathrm{i}}{\ln 2}\wedge x=\frac{\ln (u\mp \frac{\sqrt{3}(u-1)}{3}\mathrm{i})+2n_2\pi \mathrm{i}}{\ln 3}$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{find}{n}_1,n_2\in \mathbb{Z}\mathrm{for}u\ne 1$$$$p\pm q\mathrm{i}=\sqrt{p^2+q^2}{\mathrm{e}}^{\pm \mathrm{i}{\tan }^{-1}\frac{q}{p}}$$$$\ln (p\pm q\mathrm{i})=\frac{\ln (p^2+q^2)}{2}+(2n\pi \pm {\tan }^{-1}\frac{q}{p})\mathrm{i}$$$$\mathrm{we}\mathrm{have}$$$$\frac{\ln (p-q\mathrm{i})}{\ln 2}=\frac{\ln (p+q\mathrm{i})}{\ln 3}$$$$\mathrm{we}\mathrm{need}$$$$\frac{2n_1\pi -{\tan }^{-1}t}{\ln 2}=\frac{2n_2\pi +{\tan }^{-1}t}{\ln 3}$$$$\iff$$$$n_2=\frac{1-\frac{\ln 3}{\ln 2}}{2\pi }{\tan }^{-1}t+\frac{\ln 3}{\ln 2}{n}_1\mathrm{with}{n}_1,n_2\in \mathbb{Z}$$

Commented by a.lgnaoui last updated on 23/Sep/22

$$thankyou$$

Commented by Tawa11 last updated on 23/Sep/22

$$\mathrm{Great}\mathrm{sirs}$$

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