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Question Number 165433 by mathlove last updated on 01/Feb/22

Answered by aleks041103 last updated on 01/Feb/22

$$$$$$\frac{3}{2^x}+\frac{3}{3^x}=2\frac{{\left(3/2\right)}^x}{2^x}+2\frac{{\left(3/2\right)}^{-x}}{3^x}$$$$\Rightarrow \frac{3-2{\left(3/2\right)}^x}{2^x}=\frac{2{\left(3/2\right)}^{-x}-3}{3^x}$$$$\Rightarrow {\left(\frac{3}{2}\right)}^x(3-2{\left(\frac{3}{2}\right)}^x)=2{\left(\frac{3}{2}\right)}^{-x}-3$$$$t={\left(\frac{3}{2}\right)}^x$$$$\Rightarrow t(3-2t)=\frac{2}{t}-3$$$$3t^2-2t^3+3t-2=0$$$$2t^3-3t^2-3t+2=0$$$$2(t^3+1)-3t(t+1)=0$$$$2(t+1)(t^2-t+1)-3t(t+1)=0$$$$(t+1)(2t^2-5t+2)=0$$$$t_1=-1$$$$t_{2,3}=\frac{5\pm \sqrt{25-4.2.2}}{4}=\frac{5\pm 3}{4}=\frac{1}{2};2$$$$t={\left(\frac{3}{2}\right)}^x\Rightarrow x={log}_{3/2}t=\frac{{log}_{2}t}{{log}_{2}3-1}$$$$t⩾0$$$$x_1=\frac{{log}_2\frac{1}{2}}{{log}_{2}3-1}=\frac{1}{1-{log}_{2}3}$$$$x_2=\frac{{log}_{2}2}{{log}_{2}3-1}=\frac{1}{{log}_{2}3-1}$$$$$$$$x=\frac{1}{{log}_{2}3-1};\frac{1}{1-{log}_{2}3}$$

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