Question

(UFG) SE LOG 2 (X-Y)=A E X+Y=8 então log2(x²-y²)vale
a-2+a
b-3a
c-3+a
d-8
e-8+a

Answer 1

$\displaystyle \sf \left \{ {{\displaystyle log_{\ 2}\ (x-y) = a \atop {\displaystyle x+y=8}} \right. \ \ \ ; \ \ log_{\ 2} \ (x^2-y^2) = \ ? \\\\\\ \underline{Fa{\c c}amos}: \\\\ x+y = 8\\\\ log_{\ 2} (x+y) =log_{\ 2} \ 8 \\\\\ log_{\ 2} \ (x+y) = log_{\ 2} \ 2^{3} \\\\ log_{\ 2 }\ (x+y) = 3 \\\\\\ \underline{Da{\'i}}} : \\\\ log_{\ 2}\ (x-y) = a \\\\ \underline{log_{\ 2}\ (x+y) = 3 \ \ \ + } \\\\ log_{\ 2}\ (x-y)+log_{\ 2}(x+y) = 3+a$

$\displaystyle \sf log_{\ 2} \ [(x-y)(x+y)] = 3+a \\\\\\ \huge\boxed{\sf log_{\ 2}(x^2-y^2)= 3+a }\checkmark$