Question

me ajudem !!!! calcular pela definição os seguintes Logaritmos:a)log2 1sobre8 b)log8 4 c) log0,25 32

Answer 1

$a) Log_2\ \frac{1}{8}= Log_2\ 8^{-1}= Log_2\ 2^{-3}= -3*Log_2\ 2= -3*(1)= \boxed{-3} \\ \\ \boxed{Log_2\ \frac{1}{8}=-3} \\ \\ \\ b) Log_84= x \to 8^x= 4\to (2^3)^x= 2^2\to 3x=2\to x= \frac{2}{3} \\ \\ \boxed{Log_8a= \frac{2}{3}} \\ \\ \\ c) Log_ {0,25}32= x\to 0,25^x= 32\to \frac{1}{4}^x= 2^5\to (2^2) ^{-1*x}= 2^5 \to -2x= 5 \\ \\ x= - \frac{2}{5} \\ \\ \boxed{Log_ {0,25}32= - \frac{2}{5}}$