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Question Number 126161 by TheMexicanTacosG last updated on 17/Dec/20
find “a” ^a (√(a^2 )) = ((4^(−(1/2)) )^(1/2) )^(−2)
$${find}\:\:“{a}'' \\ $$$$ \\ $$$$\:^{{a}} \sqrt{{a}^{\mathrm{2}} \:}\:=\:\left(\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{2}} \:\:\:\:\: \\ $$
Answered by Olaf last updated on 18/Dec/20
(a^2 )^(1/a) = [(4^(−(1/2)) )^(1/2) ]^(−2) = (4^(−(1/2)) )^(−1) = 4^(1/2) = 2 a^(2/a) = 2 (2/a)lna = ln2 2lna = aln2 a^2 = 2^a a = 2 or 4
$$\sqrt[{{a}}]{{a}^{\mathrm{2}} }\:=\:\left[\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]^{−\mathrm{2}} \:=\:\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \:=\:\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{2} \\ $$$${a}^{\frac{\mathrm{2}}{{a}}} \:=\:\mathrm{2} \\ $$$$\frac{\mathrm{2}}{{a}}\mathrm{ln}{a}\:=\:\mathrm{ln2} \\ $$$$\mathrm{2ln}{a}\:=\:{a}\mathrm{ln2} \\ $$$${a}^{\mathrm{2}} \:=\:\mathrm{2}^{{a}} \\ $$$${a}\:=\:\mathrm{2}\:\mathrm{or}\:\mathrm{4} \\ $$
Answered by talminator2856791 last updated on 17/Dec/20
a^(2/a) = ((4^(− (1/2)) )^(1/2) )^(−2) a^(1/a) ∙ a^(1/a) = ((4^(− (1/2)) )^(1/2) )^(−1) ∙ ((4^(− (1/2)) )^(1/2) )^(−1) a^(1/a) = ((4^(− (1/2)) )^(1/2) )^(−1) a^(1/a) = ((4^(1/2) ∙ 4^(−1) )^(1/2) )^(−1) a^(1/a) = (4^(1/4) ∙ 4^(− (1/2)) )^(−1) a^(1/a) = (4^(− (1/4)) ∙ 4^( (1/2)) ) a^(1/a) = 4^(1/4) a = 4
$$\: \\ $$$$\:{a}^{\frac{\mathrm{2}}{{a}}} \:=\:\left(\left(\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{2}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:\centerdot\:{a}^{\frac{\mathrm{1}}{{a}}} \:=\:\left(\left(\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \centerdot\:\left(\left(\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:\:=\:\left(\left(\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:=\:\left(\left(\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \centerdot\:\mathrm{4}^{−\mathrm{1}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:=\:\left(\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} \centerdot\:\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{1}} \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:=\:\left(\mathrm{4}^{−\:\frac{\mathrm{1}}{\mathrm{4}}} \centerdot\:\mathrm{4}^{\:\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$\:{a}^{\frac{\mathrm{1}}{{a}}} \:=\:\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:{a}\:=\:\mathrm{4} \\ $$$$\: \\ $$$$\: \\ $$
Answered by JDamian last updated on 17/Dec/20
((4^(−(1/2)) )^(1/2) )^(−2) = 4^(1/2) = 2^(2/2) a^(2/a) = 2^(2/2) → a= { (2),(4) :}
$$\left(\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{2}} =\:\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{2}}} \\ $$$${a}^{\frac{\mathrm{2}}{{a}}} =\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{2}}} \:\:\:\rightarrow\:\:\:\:{a}=\begin{cases}{\mathrm{2}}\\{\mathrm{4}}\end{cases} \\ $$

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