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Question Number 94868 by O Predador last updated on 21/May/20

Commented by john santu last updated on 21/May/20

$${\mathrm{n}}^2=2^{2\mathrm{n}}\Rightarrow \mathrm{n}=2^{\mathrm{n}}$$$$\ln \left(\mathrm{n}\right)=\mathrm{n}\ln \left(2\right)$$$$\frac{\ln \left(\mathrm{n}\right)}{\mathrm{n}}=\ln \left(2\right)\Rightarrow \ln {\left(\mathrm{n}\right)}^{\frac{1}{\mathrm{n}}}=\ln \left(2\right)$$$${\mathrm{n}}^{\frac{1}{\mathrm{n}}}=2;\mathrm{we}\mathrm{can}\mathrm{aply}\mathrm{Lambert}$$$$\mathrm{W}\mathrm{function}$$

Commented by hknkrc46 last updated on 21/May/20

$${\mathrm{n}}^2=4^{\mathrm{n}}\Rightarrow {\mathrm{n}}^2={\left(2^2\right)}^{\mathrm{n}}\Rightarrow {\mathrm{n}}^2={\left(2^{\mathrm{n}}\right)}^2\Rightarrow \mathrm{n}=2^{\mathrm{n}}$$$$$$

Commented by O Predador last updated on 21/May/20

$$$$$$\mathrm{Obrigado}!$$

Commented by mr W last updated on 21/May/20

$$forn>0:4^n={\left(2^n\right)}^2>n^2$$$$therefore{2}^n=4^{n}hasnosolution!$$

Commented by hknkrc46 last updated on 21/May/20

$$\mathrm{De}\mathrm{nada}!$$

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