Question

se (a-b)= 16 e (a+b)=8, então, log₂(a²-b²) e igual a?

Answer 1

(a² - b²) = (a - b)(a + b)

$log_{2}( a^{2} - b^{2} )= \\ \\ log_{2}(a+b)(a-b)$

temos:

(a - b) = 16

(a + b) = 8

$log_{2} (16.8)= \\ \\ log_{2}128= \\ \\ log_{2} 2^{7}= \\ \\ 7. log_{2}2= \\ \\ 7.1= \\ \\ 7$