Question

sendo A= 1024/ raiz cubica de 256 e B= 1/64 o valor de log na base 2 ^a+ log na base 2 ^b

Answer 1

$a = 1024/\sqrt[3]{256}$
$a=2^{10}/\sqrt[3]{2^{8}}$
$a=2^{10}/(2^{8/3})$
$a=2^{(10 - [8/3])}$
$a=2^{(30-8)/3)$
$a=2^{22/3}$

$b=1/64$
$b=1/2^{6}$
$b=2^{-6}$

$log_{2}(a) + log_{2}(b) = log_{2}(a*b)$
$log_{2}(a) + log_{2}(b) = log_{2}(2^{22/3}*2^{-6})$
$log_{2}(a) + log_{2}(b) = log_{2}(2^{([22/3] - 6)})$
$log_{2}(a) + log_{2}(b) = log_{2}(2^{([22-18]/3)})$
$log_{2}(a) + log_{2}(b) = log_{2}(2^{(4/3)})$
$log_{2}(a) + log_{2}(b) = (4/3)*log_{2}(2)$
$log_{2}(a) + log_{2}(b) = (4/3)*1$
$log_{2}(a) + log_{2}(b) = 4/3$