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Question Number 132574 by liberty last updated on 15/Feb/21
$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{log}\:_{\mathrm{2020}} \left(\mathrm{x}\right)\:\mathrm{and}\: \\ $$$$\mathrm{p}^{\left(\mathrm{p}\right)^{\mathrm{p}^{\mathrm{2020}} } } \:=\:\mathrm{2020}\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{p}\right)\:=\:... \\ $$$$\left(\mathrm{a}\right)\sqrt[{\mathrm{2020}}]{\mathrm{2020}}\:\:\:\:\:\left(\mathrm{c}\right)\:\sqrt[{\mathrm{2020}}]{\frac{\mathrm{1}}{\mathrm{2020}}} \\ $$$$\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{2020}}\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{2020}\:\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2020}\right) \\ $$
Answered by mr W last updated on 15/Feb/21
$${p}^{{p}^{{p}^{\mathrm{2020}} } } =\mathrm{2020} \\ $$$$\Rightarrow{p}=\mathrm{2020}^{\frac{\mathrm{1}}{\mathrm{2020}}} \\ $$$${f}\left({p}\right)=\mathrm{log}_{\mathrm{2020}} \:\mathrm{2020}^{\frac{\mathrm{1}}{\mathrm{2020}}} =\frac{\mathrm{1}}{\mathrm{2020}} \\ $$$$\Rightarrow\left({b}\right) \\ $$
Commented by liberty last updated on 15/Feb/21
$$\mathrm{waw}.... \\ $$