Question

a solução da equação log2 0,5+ log2 x - log2 raiz quadrada de 2= 2 está contida no intervalo

Answer 1

$\displaystyle \log_20,5+\log_2x+co\log_2\sqrt{2}=2\\\\\log_2\left(\frac{1}{2}\right)=\log_21-\log_22=-1\\\\co\log_2\sqrt{2}=-\log_22^{\frac{1}{2}}=-\frac{1}{2}\log_22=-\frac{1}{2}\cdot1=-\frac{1}{2}\\\\-1+\log_2x-\frac{1}{2}=2\implies \log_2x=2+1+\frac{1}{2}\implies \\\\\log_2x=\frac{4}{2}+\frac{2}{2}+\frac{1}{2}\implies \log_2x=\frac{7}{2}\Longleftrightarrow 2^{\frac{7}{2}}=x\implies \\\\\sqrt{2^7}=x\implies \sqrt{128}=x\implies x=\sqrt{2^22^22^22}\implies x=2.2.2\sqrt2\\\\\boxed{x=8\sqrt2}$