Question

Sabendo que a e b são numeris reais que satisfazem a igualdade 2log (a-2b) - log a - log b = 0, calcule o valor de a/b. Justifique detalhadamente sua resposta.

Answer 1

$\displaystyle \sf \text{Fa{\c c}amos a condi{\c c}{\~a}o de exist{\^e}ncia do log} : \\\\ 2\cdot \log(\alpha- 2\beta) \to \alpha -2\beta > 0 \to \boxed{\sf \beta < \frac{\alpha}{2} } \\\\ \log \alpha \to \boxed{\sf \alpha > 0 }\\\\ \log \beta \to \boxed{\sf \beta > 0 }$

Desenvolvendo a equação, aplicando as propriedades de logaritmos:

$\displaystyle \sf 2\cdot \log(\alpha-2\beta) -\log\alpha -\log\beta = 0 \\\\ 2\cdot \log(\alpha -2\beta ) = \log \alpha +\log \beta \\\\ \log(\alpha-2\beta )^2 = \log (\alpha\cdot \beta ) \\\\ (\alpha -2\beta )^2=\alpha \cdot \beta \\\\ \alpha ^2-4\cdot \alpha\cdot \beta +4\cdot \beta^2= \alpha\cdot \beta \\\\ \alpha ^2-4\cdot\alpha\cdot \beta -\alpha\cdot \beta +4\cdot \beta ^2=0 \\\\ (\alpha)^2-5\cdot (\alpha)\cdot \beta +4\cdot \beta^2= 0 \\\\ \underline{\text{Resolvendo para }(\alpha) \ temos }:$

$\displaystyle \sf \alpha = \frac{-(-5\cdot \beta )\pm\sqrt{(-5\cdot \beta )^2-4\cdot 1\cdot (4\cdot \beta ^2)}}{2\cdot 1} \\\\\\ \alpha = \frac{5\cdot \beta \pm\sqrt{25\cdot \beta ^2-16\cdot \beta ^2 }}{2} \to \alpha = \frac{5\cdot \beta \pm\sqrt{9\cdot \beta^2 }}{2}$

$\displaystyle \sf \left\{ \begin{array}{I} \displaystyle \sf \alpha = \frac{5\cdot \beta +3\cdot \beta }{2}\to \boxed{\sf \alpha = 4\cdot \beta } \\\\\\ \displaystyle \sf \alpha = \frac{5\cdot \beta -3\cdot \beta}{2} \to \boxed{\sf \alpha = \beta }\ \end{array} \right \\\\\\ \alpha =\beta \ (\text{N{\~A}O CONV{\'EM}} ), \ pois \ \boxed{\beta < \frac{\alpha}{2} }$

Portanto :

$\alpha = 4\beta$

$\displaystyle \sf \text{Dividindo ambos os lados por }\beta : \\\\ \frac{\alpha }{\beta } = 4 \\\\\ Portanto : \\\\ \huge\boxed{\frac{\alpha }{\beta }=4 \ }\checkmark$