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Question Number 173485 by mnjuly1970 last updated on 12/Jul/22

$$$$$$Solve(x\in \mathbb{R})$$$$x{2}^{\frac{1}{x}}+\frac{1}{x}{2}^x=4$$$$-Source:L.Panaitopol$$

Answered by aleks041103 last updated on 12/Jul/22

$$obvx>0forx{2}^{1/x}+2^x/x=4$$$$f\left(x\right)=x{2}^{1/x}$$$$\Rightarrow g\left(x\right)=f\left(x\right)+f\left(1/x\right)$$$$\Rightarrow g\left(1/x\right)=f\left(1/x\right)+f\left(x\right)$$$$\Rightarrow g\left(x\right)=g\left(1/x\right)=4$$$$\Rightarrow thereforeitisenoughtosearch$$$$forx>1andfor\mathrm{\forall }{x}_i>1,s.t.g\left(x_i\right)=4{we}^{\prime }ll$$$$have{\overline{x}}_i=\frac{1}{x_i},s.t.g\left({\overline{x}}_i\right)=4too.$$$$2^x=e^{ln\left(2\right)x}=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)x\right)}^i}{i!}$$$$\Rightarrow f\left(x\right)=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)\right)}^i}{i!}\frac{1}{x^{i-1}}$$$$\Rightarrow f\left(1/x\right)=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)\right)}^i}{i!}{x}^{i-1}$$$$\Rightarrow g\left(x\right)=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)\right)}^i}{i!}(x^{i-1}+\frac{1}{x^{i-1}})$$$$g\left(x\right)=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)\right)}^i}{i!}{s}_{i-1}\left(x\right)$$$$s_i\left(x\right)=x^i+x^{-i}$$$$s_i^{\prime }\left(x\right)={ix}^{i-1}-{ix}^{-i-1}=\frac{i}{x}(x^i-x^{-i})$$$$forx>1:$$$$i>0:x^i>1\Rightarrow x^{-i}<1\Rightarrow x^i-x^{-i}>0$$$$\Rightarrow s_{i>0}^{\prime }(x>1)>0$$$$i<0:x^{i<1}\Rightarrow x^{-i}>1\Rightarrow x^i-x^{-i}<0$$$$\Rightarrow i(x^i-x^{-i})>0$$$$\Rightarrow s_{i<0}^{\prime }(x>1)>0$$$$i=0:$$$$s_0^{\prime }=0$$$$\Rightarrow g^{\prime }(x>1)=\underset{i=0,i\ne 1}{\sum ^{\mathrm{\infty }}}\frac{{\left(ln\left(2\right)\right)}^i}{i!}{s}_{i-1}^{\prime }(x>1)$$$$since{s}_{i-1}^{\prime }(x>1)>0\Rightarrow g^{\prime }(x>1)>0$$$$\Rightarrow g\left(x\right)isstrictlyincreasingforx>1$$$$\Rightarrow forg\left(x\right)=4existsatmost1soln.x⩾1$$$$Andg\left(1\right)=4$$$$\Rightarrow Theonlysolutiontotheproblemis$$$$x=1$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by dragan91 last updated on 12/Jul/22

$$2^{{\log }_2\mathrm{x}+\frac{1}{\mathrm{x}}}+2^{{\log }_2\frac{1}{\mathrm{x}}+\mathrm{x}}=4$$$$2^{{\log }_2\mathrm{x}+\frac{1}{\mathrm{x}}}+2^{-{\log }_2\mathrm{x}+\mathrm{x}}=4$$$$2^{{\log }_2\mathrm{x}+\frac{1}{\mathrm{x}}}+2^{-{\log }_2\mathrm{x}+\mathrm{x}}\stackrel{\mathrm{AM}-\mathrm{GM}}{⩾}2\sqrt{2^{\mathrm{x}+\frac{1}{\mathrm{x}}}}$$$$2^{\mathrm{x}+\frac{1}{\mathrm{x}}}⩽2^2\mid {\log }_2$$$$\mathrm{x}+\frac{1}{\mathrm{x}}⩽2$$$$\mathrm{since}\mathrm{x}+\frac{1}{\mathrm{x}}\stackrel{\mathrm{AM}-\mathrm{GM}}{⩾}2\Rightarrow \mathrm{equation}\mathrm{has}\mathrm{solution}$$$$\mathrm{only}\mathrm{for}\mathrm{x}=1$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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