Question

Calcule os logaritmos: log$log_{3}$ 27 / log_{√2} 64 / log_{0,2} 25 / log_{√13} \sqrt{13} / log_{\frac{3}{2}} \frac{9}{4} / log_{3} √\frac{1}{27}

Answer 1

$a) \\ log_{3} 27 =x \\ 3^{x}=27 \\ 3^{x}=3^{3} \\ x=3$

$b) \\ log_{ \sqrt{2} } 64=x \\ (2^{ \frac{1}{2} })^{x}=64 \\ 2^{ \frac{x}{2} }=2^{6} \\ \frac{x}{2}=6 \\ x=2.6 \\ x=12$

$c) \\ log_{0,2} 25=x \\ (0,2)^{x}=25 \\ ( \frac{1}{5})^{x}=25 \\ 5^{-x}=5^{2} \\ -x=2 \\ x=-2$

$d) \\ log_{ \sqrt{13} } \sqrt{13} =x \\ (13^{ \frac{1}{2} })^{x} = 13^{ \frac{1}{2} } \\ 13^{ \frac{x}{2} }=13^{ \frac{1}{2} } \\ \frac{x}{2}= \frac{1}{2} \\ x= \frac{2}{2} \\ x=1$

$e) \\ log_{ \frac{3}{2} } \frac{9}{4} =x \\ ( \frac{3}{2} )^{x} = \frac{9}{4} \\ ( \frac{3}{2})^{x} = (\frac{3}{2})^{2} \\ x=2$

$f) \\ log_{3} \frac{1}{27}=x \\ 3^{x} = \frac{1}{27} \\ 3^{x}= \frac{1}{ 3^{3} } \\ 3^{x} = 3^{-3} \\ x=-3$

Espero ajudar... (: