[tex]\displaystyle \sf \text{a \'e um n\'umero maior que 0 pois se trata de uma exponencial, ent\~ao}: \\\\ a = \frac{\sqrt{29}-1}{2} \\\\\\ 2^{x^{\frac{1}{2}}} = \frac{\sqrt{29}-1}{2} \\\\\\ \log_22^{x^{\frac{1}{2}}} = \log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ x^{\frac{1}{2}}\cdot \underbrace{\sf \log_22}_{1} = \log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ x^{\frac{1}{2}} =\log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ \text{Elevando ambos os lados ao quadrado} :[/tex]
[tex]\Large\boxed{\sf \ x = \left[\ \log_2\left(\frac{\sqrt{29}-1}{2}\right) \ \right]^2\ }\checkmark[/tex]
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x é aproximadamente 1,283
Verified answer
[tex]\displaystyle \sf 4^{x^{\frac{1}{2}}} +2^{x^{\frac{1}{2}}} = 7 \\\\ (2^2)^{x^{\frac{1}{2}}} +2^{x^{\frac{1}{2}}} ^= 7 \\\\ 2^{2\cdot (x^{\frac{1}{2}})} +2^{x^{\frac{1}{2}}} = 7 \\\\ \text{Fa\c camos} : \\\\ 2^{x^{\frac{1}{2}}}= a \to 2^{2\cdot (x^{\frac{1}{2}})} = a^2 \\\\ Da{\'i}}: \\\\ a^2+a = 7 \\\\ a^2+a-7 = 0 \\\\ a = \frac{-1\pm\sqrt{1^2-4\cdot 1\cdot(-7)}}{2} \\\\\\ a = \frac{-1\pm\sqrt{1+28}}{2} \\\\\\ a = \frac{-1\pm\sqrt{29}}{2}[/tex]
[tex]\displaystyle \sf \text{a \'e um n\'umero maior que 0 pois se trata de uma exponencial, ent\~ao}: \\\\ a = \frac{\sqrt{29}-1}{2} \\\\\\ 2^{x^{\frac{1}{2}}} = \frac{\sqrt{29}-1}{2} \\\\\\ \log_22^{x^{\frac{1}{2}}} = \log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ x^{\frac{1}{2}}\cdot \underbrace{\sf \log_22}_{1} = \log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ x^{\frac{1}{2}} =\log_2\left(\frac{\sqrt{29}-1}{2}\right) \\\\\\ \text{Elevando ambos os lados ao quadrado} :[/tex]
[tex]\Large\boxed{\sf \ x = \left[\ \log_2\left(\frac{\sqrt{29}-1}{2}\right) \ \right]^2\ }\checkmark[/tex]