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Question Number 138579 by Jamshidbek last updated on 15/Apr/21

Answered by MJS_new last updated on 16/Apr/21

x≈5.40343170

$${x}\approx\mathrm{5}.\mathrm{40343170} \\ $$

Commented by MJS_new last updated on 16/Apr/21

we can only approximate  x^x^x  =2017^(2017)   x^x ln x −2017ln 2017 =0  now use a calculator

$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}^{{x}^{{x}} } =\mathrm{2017}^{\mathrm{2017}} \\ $$$${x}^{{x}} \mathrm{ln}\:{x}\:−\mathrm{2017ln}\:\mathrm{2017}\:=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{a}\:\mathrm{calculator} \\ $$

Commented by Jamshidbek last updated on 16/Apr/21

I need solution

$$\mathrm{I}\:\mathrm{need}\:\mathrm{solution} \\ $$

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