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Question Number 194869 by sonukgindia last updated on 17/Jul/23

Answered by Frix last updated on 17/Jul/23

$$-x^{2^{-x^4}}=-x^{4^{-x^2}}$$$$x^{{\left(1/2\right)}^{x^4}}=x^{{\left(1/4\right)}^{x^2}}$$$${\left(1/2\right)}^{x^4}\ln x={\left(1/4\right)}^{x^2}\ln x\Rightarrow x=1$$$${\left(1/2\right)}^{x^4}={\left(1/4\right)}^{x^2}$$$$x^4\ln \frac{1}{2}=x^2\ln \frac{1}{4}$$$$-x^4\ln 2=-2x^2\ln 2$$$$x^4=2x^2\Rightarrow x=0$$$$x^2=2\Rightarrow x=\pm \sqrt{2}$$

Commented by Frix last updated on 18/Jul/23

$$-x^{2^{-x^4}}=-(x\uparrow (2\uparrow (-(x\uparrow 4))))$$$$-x^{4^{-x^2}}=-(x\uparrow (4\uparrow (-(x\uparrow 2))))$$$$\mathrm{Both}\mathrm{are}\mathrm{identical}\mathrm{for}x=-\sqrt{2}$$$$-{(-\sqrt{2})}^4=-42^{-4}=\frac{1}{16}-{(-\sqrt{2})}^{\frac{1}{16}}=-2^{\frac{1}{32}}\mathrm{i}$$$$-{(-\sqrt{2})}^2=-24^{-2}=\frac{1}{16}-{(-\sqrt{2})}^{\frac{1}{16}}=-2^{\frac{1}{32}}\mathrm{i}$$$$\mathrm{It}\mathrm{says}‘‘x\in {\mathbb{R}}^{″}\mathrm{and}-\sqrt{2}\in \mathbb{R}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{like}$$$$\mathrm{Solve}\mathrm{for}x\in \mathbb{R}:{(1+4\mathrm{i})}^x={(3-2\mathrm{i})}^x\Rightarrow x=0\in \mathbb{R}$$

Commented by aba last updated on 18/Jul/23

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