Dissertativa 6 - 2ª Fase - Dia 2 - Unicamp 2026

a) Para k = 1, b = 2 e d = 4:

$\left\{\begin{array}{l}({\log }_2 2) \mathrm{x} + ({\log }_4 2) \mathrm{y} = 1\\ ({\log }_4 4) \mathrm{x} + ({\log }_2 4) \mathrm{y} = 0\end{array}\right.$

Resolvendo os logaritimos:

$\left\{\begin{array}{l}\mathrm{x} + \frac{1}{2} \mathrm{y} = 1\\ \mathrm{x} + 2\mathrm{y} = 0\end{array}\right.$

Assim, $\mathrm{x} = \frac{4}{3}$ e $\mathrm{y} = -\frac{2}{3}$

b) Para k = 0, obtém-se o sistema homogêneo:

$\left\{\begin{array}{l}({\log }_{\mathrm{b}} 2) \mathrm{x} + ({\log }_{4 } \mathrm{b}) \mathrm{y} = 0\\ ({\log }_{\mathrm{d}} 4) \mathrm{x} + ({\log }_2 \mathrm{d}) \mathrm{y} = 0\end{array}\right.$

Para que um sistema homogêneo admita infinitas soluções:

$$\begin{gathered} \left|\begin{array}{cc}{\log }_{\mathrm{b}} 2& {\log }_4 \mathrm{b}\\ {\log }_{\mathrm{d}} 4& {\log }_2 \mathrm{d}\end{array}\right| = 0 \\ {\log }_{\mathrm{b}} 2 · {\log }_2 \mathrm{d} - {\log }_{\mathrm{d}} 4 · {\log }_4 \mathrm{b} = 0 \end{gathered}$$

Aplicando a mudança de base e escolhendo a base b, tem-se:

$$\begin{gathered} {\log }_b 2 · \frac{{\log }_b d}{{\log }_b 2}- \frac{{\log }_b 4}{{\log }_b d} · \frac{{\log }_b b}{{\log }_b 4} = 0 \\ {\log }_{\mathrm{b}} \mathrm{d} - \frac{1}{{\log }_{\mathrm{b}}\mathrm{d}} = 0 \\ {\log }_{\mathrm{b}} \mathrm{d} = \frac{1}{{\log }_{\mathrm{b}} \mathrm{d}} \\ {\left({\log }_{\mathrm{b}} \mathrm{d}\right)}^2 = 1 \\ {\log }_{\mathrm{b}} \mathrm{d} = \pm 1 \end{gathered}$$

Assim, deve-se ter $\mathrm{b} = \mathrm{d}$ ou $\mathrm{b} = \frac{1}{\mathrm{d}}$ para que o sistema admita infinitas soluções.