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Question Number 126161 by TheMexicanTacosG last updated on 17/Dec/20

$$find‘‘a^{″}$$$$$$$${}^a\sqrt{a^2}={\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-2}$$

Answered by Olaf last updated on 18/Dec/20

$$\sqrt[a]{a^2}={\left[{\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right]}^{-2}={\left(4^{-\frac{1}{2}}\right)}^{-1}=4^{\frac{1}{2}}=2$$$$a^{\frac{2}{a}}=2$$$$\frac{2}{a}\ln a=\ln 2$$$$2\ln a=a\ln 2$$$$a^2=2^a$$$$a=2\mathrm{or}4$$

Answered by talminator2856791 last updated on 17/Dec/20

$$$$$$a^{\frac{2}{a}}={\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-2}$$$$a^{\frac{1}{a}}\cdot a^{\frac{1}{a}}={\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-1}\cdot {\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-1}$$$$a^{\frac{1}{a}}={\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-1}$$$$a^{\frac{1}{a}}={\left({(4^{\frac{1}{2}}\cdot 4^{-1})}^{\frac{1}{2}}\right)}^{-1}$$$$a^{\frac{1}{a}}={(4^{\frac{1}{4}}\cdot 4^{-\frac{1}{2}})}^{-1}$$$$a^{\frac{1}{a}}=(4^{-\frac{1}{4}}\cdot 4^{\frac{1}{2}})$$$$a^{\frac{1}{a}}=4^{\frac{1}{4}}$$$$a=4$$$$$$$$$$

Answered by JDamian last updated on 17/Dec/20

$${\left({\left(4^{-\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{-2}=4^{\frac{1}{2}}=2^{\frac{2}{2}}$$$$a^{\frac{2}{a}}=2^{\frac{2}{2}}\to a=\{\begin{array}{l}2\\ 4\end{array}$$

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