Question

Necesito ayuda con este problema :c

Answer 1

$\mathrm{\large{Hallar:\ 2^{A+B}}}$

$\begin{cases}A=log_{2}(m^2-n^2)-log_{2}(m+n)\cr \cr B=\frac{log_{2}(m^2-2mn+n^2)}{log_{2}(4)}\end{cases}\Rightarrow m > n$

$\mathrm{\large{Productos\ notables:}}$

$m^2-n^2=(m+n)(m-n)$

$m^2-2mn+n^2=(m-n)^2$

$\mathrm{\large{Propiedades \ de\ los\ logaritmos:}}$

$log_a (b.c)=log_a (b)+log_a(c)\\ log_a(\frac{b}{c})=log_a (b)-log_a (c)\ si\ c≠0\\ log_a(b^c)=c.log_a(b)\\ a^{log_a (b)}=b$

$\mathrm{\large{Resolviendo:}}$

$A=log_{2}\left[(m+n)(m-n)\right]-log(m+n)\\ A=log_{2}(m+n)+log_{2}(m-n)-log_{2}(m+n)\\ A=log_{2}(m-n)$

$B=\frac{log\left[(m-n)^2\right]}{log_{2}4}\\ B=\frac{2\left[log_{2}(m-n)\right]}{2}\\ B=log_{2}(m-n)$

$\mathrm{\large{Entonces:}}$

$2^{A+B}\Rightarrow 2^{log_{2}(m-n)+log_{2}(m-n)}=2^{2log_{2}(m-n)}\\ 2^{log_{2}(m-n)^2}\Rightarrow (m-n)^2$

$\mathrm{\large{Respuesta:c)\ (m-n)^2}}$

$\mathrm{\large{PD:}}$

$log_{2}(4)=2$