Question

Log½32
calculo dessa operação

Answer 1

Pela definiçao de logaritmo temos que:

$log_{ \frac{1}{2} } 32 =x\\ \\32= ( \frac{1}{2})^{x} \\ \\ 2^{5}= ( \frac{1}{2})^{x} \\ \\ 2^{5}= 2^{-x} \\ \\ -x=5 \\ \\ x=-5$

Answer 2

Log(base 1/2) 32 = X
log(base 1/2) 2^5 = X
\frac{1}{2} ^x =2^5

2^-x=2^5

X=-5