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Question Number 71824 by mr W last updated on 20/Oct/19

$$solve$$$$x^{x^{x^{2019}}}=2019$$

Commented by mr W last updated on 21/Oct/19

$$wecansee:$$$$ifweassume{x}^{2019}=2019,wewillget$$$$x^{x^{2019}}=x^{2019}=2019$$$$x^{x^{x^{2019}}}=x^{2019}=2019$$$$x^{x^{x^{x^{2019}}}}=x^{2019}=2019$$$$....$$$$i.e.theassumption{x}^{2019}=2019iscorrect.$$$$from{x}^{2019}=2019,weget$$$$\Rightarrow x=\sqrt[2019]{2019}$$

Answered by MJS last updated on 20/Oct/19

$$\mathrm{usually}{x}^{x^{x^{2019}}}=x^{\left(x^{\left(x^{2019}\right)}\right)}$$$$x^{\left(x^{\left(x^{2019}\right)}\right)}=2019\Rightarrow$$$$\Rightarrow 1.00377648280<x<1.00377648281$$$$$$$$\mathrm{but}\mathrm{if}\mathrm{you}\mathrm{mean}$$$${\left({\left(x^x\right)}^x\right)}^{2019}=2019\Rightarrow$$$$\Rightarrow 1.00374827787<x<1.00374827788$$

Commented by mr W last updated on 21/Oct/19

$$thankssir!$$$$thefirstinterpretationismeant.$$$$theexactvalueisx=\sqrt[2019]{2019}$$

Commented by TawaTawa last updated on 21/Oct/19

$$\mathrm{Please}\mathrm{workings}\mathrm{sir}$$

Commented by mr W last updated on 22/Oct/19

$$seeabove.$$$$theresultisthesame,nomatter$$$$howmanytimesxoccursinthe$$$$equation.$$

Answered by mind is power last updated on 23/Oct/19

$${\mathrm{x}}^{{\mathrm{x}}^{2019}}\ln \left(\mathrm{x}\right)=\ln \left(2019\right)$$$$\Rightarrow {\mathrm{x}}^{2019}\ln \left(\mathrm{x}\right)+\ln \left(\ln \left(\mathrm{x}\right)\right)=\ln \left(\ln \left(2019\right)\right)$$$$\Rightarrow {\mathrm{x}}^{2019}\ln \left(\mathrm{x}\right)=\ln \left(\frac{\ln \left(2019\right)}{\ln \left(\mathrm{x}\right)}\right)$$$$\mathrm{u}=\frac{\ln \left(2019\right)}{\ln \left(\mathrm{x}\right)}\Rightarrow \ln \left(\mathrm{x}\right)=\frac{\ln \left(2019\right)}{\mathrm{u}}=\Rightarrow \mathrm{x}={2019}^{\frac{1}{\mathrm{u}}}$$$$\Rightarrow {2019}^{\frac{2019}{\mathrm{u}}}.\frac{\ln \left(2019\right)}{\mathrm{u}}=\ln \left(\mathrm{u}\right)$$$$\Rightarrow {2019}^{{2019}^{\frac{2019}{\mathrm{u}}}}={\mathrm{u}}^{{\mathrm{u}}^1}=\mathrm{S}$$$$1=\frac{\mathrm{u}}{\mathrm{u}}$$$$\Rightarrow \mathrm{S}\iff {2019}^{{2019}^{\frac{2019}{\mathrm{u}}}}={\mathrm{u}}^{{\mathrm{u}}^{\frac{\mathrm{u}}{\mathrm{u}}}}\Rightarrow \mathrm{u}=2019$$$$\ln \left(\mathrm{x}\right)=\frac{\ln \left(2019\right)}{2019}\Rightarrow \mathrm{x}=\sqrt[2019]{2019}$$$$$$$$$$$$$$

Commented by mr W last updated on 23/Oct/19

$$thanksalotsir!$$

Commented by mind is power last updated on 23/Oct/19

$$\mathrm{y}{}^{\prime }\mathrm{re}\mathrm{welcom}$$

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