Question

Resolva o sistema de Logaritmo abaixo:
a) Log2 (x+y) = 2
Log5 (x-y) = 1

b) Log y (x+y) = 1
Log 3x (x+y) = 1

Answer 1

a)

$log_2(x+y)=2\\ log_5(x-y)=1\\\\ log_2(x+y)=2\\ 2^2=x+y\\ x+y=4\\\\ log_5(x-y)=1\\ 5^1=x-y\\ x-y=5\\\\ \left \{ {{x+y=4} \atop {x-y=5}} \right.\\\\ x+y+x-y=4+5\\ 2x=9\ --\ \textgreater \ \ \| \ x=\frac{9}{2}\ \|\\\\ x+y=4\\ \frac{9}{2}+y=4\\ y=4-\frac{9}{2}\\ y=\frac{8-9}{2}\ --\ \textgreater \ \ \| \ y=-\frac{1}{2}\ \| \\\\ S= \left \{ {{y=-\frac{1}{2}} \atop {x=\frac{9}{2}}} \right.$

b)

$log_y(x+y)=1\\ log_{3x}(x+y)=1\\\\ log_y(x+y)=1\\ y^1=x+y\\ x=y-y\\ \| \ x=0\ \|\\\\ log_{3x}(x+y)=1\\ (3x)^1=x+y\\ 3x-x=y\\ y=2x\\ y=2.0\\ \| \ y=0\ \|\\\\ S= \left \{ {{y=0} \atop {x=0}} \right.$