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Question Number 126484 by Mathgreat last updated on 20/Dec/20

Commented by MJS_new last updated on 20/Dec/20

$$\approx 1.63507847$$

Commented by Mathgreat last updated on 20/Dec/20

$$yechim.$$$$$$

Commented by Mathgreat last updated on 20/Dec/20

$$prove$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{just}\mathrm{use}\mathrm{a}\mathrm{calculator}$$$$f\left(x\right)=x^{x^x}-3$$$$f\left(1\right)=-2$$$$f\left(2\right)=16$$$$\Rightarrow 1<x<2$$$$f(1.5)\approx -.89$$$$\Rightarrow 1.5<x<2$$$$f(1.6)\approx -.29$$$$\Rightarrow 1.6<x<2$$$$f(1.7)\approx .70$$$$\Rightarrow 1.6<x<1.7$$$$...$$

Commented by MJS_new last updated on 20/Dec/20

$$x^{x^3}=3\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{same}\mathrm{as}{x}^{x^x}=3$$

Commented by MathSh last updated on 20/Dec/20

$$x^{x^3}=3$$$${log}_x{x}^{x^3}={log}_{x}3$$$${\left({log}_{x}3\right)}^{\left({log}_{x}3\right)}=3^3$$$$x^3=3$$$$x=\sqrt[3]{3}$$

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