Question

Resolva a equação e a inequação a
seguir.
a) log9 x + log27 x − log3 x = −1

b) 2log^(x−4) > 1

Answer 1

$a) \ log_9 \ x + log_{27} \ x - log_3 \ x = -1\\\\ \ \ \ \ \ \frac{log_3 \ x}{log_3 \ 9} + \frac{log_{3} \ x}{log_3 \ 27} - \frac{log_3 \ x}{log_3 \ 3} = -1\\\\\frac{log_3 \ x}{log_3 \ 3^2} + \frac{log_{3} \ x}{log_3 \ 3^3} - \frac{log_3 \ x}{log_3 \ 3} = -1\\\\\frac{log_3 \ x}{2\cdot log_3 \ 3} + \frac{log_{3} \ x}{3 \cdot log_3 \ 3} - \frac{log_3 \ x}{1} = -1\\\\\frac{log_3 \ x}{2\cdot 1} + \frac{log_{3} \ x}{3 \cdot 1} - \frac{log_3 \ x}{1} = -1$

$\\\frac{log_3 \ x}{3} + \frac{log_{3} \x}{3}-log_3 \ x=-1 \ \\\\log_3 \ x \cdot \Big(\frac{1}{2} + \frac{1}{3}-1\Big)=-1 \ \\\\log_3 \ x \cdot \Big(\frac{3}{6} + \frac{2}{6}-\frac{6}{6}\Big)=-1 \ \\\\log_3 \ x \cdot \Big(\frac{3+2-6}{6} \Big)=-1 \\\\log_3 \ x \cdot \Big(\frac{-1}{6} \Big)=-1 \\\\\\log_3 \ x=\frac{-1}{-1/6} \\\\log_3 \ x=-1\times \Big(-\frac{6}{1}\Big)\\\\log_3 \ x=6\\\\x=3^6$

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