Question

Use a definição para calcular :
a)log2 0,25
b)log4 8raiz2

Answer 1

Definição de logaritmo

$a^x=b \\ \\ log_{a} b=x \\ \\ a) \\ log_{2} 0,25=x \\ \\ log_{2} ( \frac{1}{4}) =x \\ \\ 2^{x} = \frac{1}{4} \\ \\ 2^{x} = \frac{1}{2^2} \\ \\ 2^{x} = 2^{-2} \\ \\ x=-2 \\ \\ \\ b) \\ log_{4} 8 \sqrt{2} =x \\ \\ 4^{x} =8 \sqrt{2} \\ \\ 2^{ 2^{x} } = 2^{3} . 2^{ \frac{1}{2} } \\ \\ 2^{2x} = 2^{ \frac{7}{2} } \\ \\ 2x= \frac{7}{2} \\ \\ x= \frac{7}{4} =1,75$

Answer 2

$a) \ log_2 \ 0,25\\ \\ =log_2 \ (\frac{25^{:25}}{100_{:25}})\\ \\=log_2 \ (\frac{1}{4}) \\ \\ = log_2 \ 2^{-2}\\ \\=-2 \cdot log_2 \ 2\\ \\=-2 \cdot 1\\ \\=-2$



$b) \ log_4 \ 8 \sqrt{2} \\ \\= \frac{log_2 \ 8 \sqrt{2} }{log_2 \ 4} \\ \\=\frac{log_2 \ 2^3\cdot 2^{ \frac{1}{2} } }{log_2 \ 2^2}\\ \\=\frac{log_2 \ 2^{3+ \frac{1}{2} } }{log_2 \ 2^2}\\ \\=\frac{log_2 \ 2^{\frac{7}{2} } }{2\cdotlog_2 \ 2}\\ \\=\frac{\frac{7}{2}log_2 \ 2 }{2 \cdot log_2 \ 2}\\ \\=\frac{\frac{7}{2} \cdot 1 }{2 \cdot 1}\\ \\=\frac{7/2}{2}\\ \\= \frac{7}{2}\cdot \frac{1}{2}\\ \\=\frac{7}{4}$