Question

Resolva o sistema.
log3 (x+y)=2
log5 (3x-y)= log5    10

Answer 1

$log_{b}(a)=c <=> b^{c}=a$
$log_{b}(a)=log_{b}(c) <=> a = c$
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$log_{3}(x+y)=2$
$3^{2}=x+y$
$9=x+y$

$log_{5}(3x-y)=log_{5}(10)$
$3x-y=10$
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$\left \{ {{x+y=9} \atop {3x-y=10}} \right.$

Somando as equações:

$x + 3x + y - y = 9 + 10$
$4x = 19$
$x = 19/4$

$x + y = 2$
$(19/4)+y=9$
$4*(19/4) + 4*y = 4*9$
$19 + 4y = 36$
$4y = 36 - 19$
$4y = 17$
$y = 17 / 4$

Answer 2

Para este problema, é primordial conhecer as seguintes propriedades logarítmicas básicas:

$\circ \ \text{Para todo log}_a b = x \Leftrightarrow a^x = b \\\\ \circ \text{Se log}_a x = \text{log}_a y, \text{ent}\tilde{a}\text{o} \ x = y$

Aplicando:

$\text{Simplificando o sistema:} \\\\ \left\{ {{\text{log}_3(x + y) \ = \ 2} \atop {\text{log}_5(3x - y) \ = \ \text{log}_5 10}}\right \\\\ \left\{ {{x + y \ = \ 3^2} \atop {3x - y \ = \ 10}}\right \\\\ \left\{ {{x + y \ = \ 9} \atop {3x - y \ = \ 10}}\right \\\\ \text{Resolvendo o sistema:} \\\\ \circ x + y = 9 \\ x = 9 - y \\ x = 9 - \frac{17}{4} \\ \boxed{x = \frac{19}{4} = 4,75} \\\\ \circ 3 \ (9 - y) - y = 10 \\ 27 - 3y - y = 10 \\ -4y = -17 \\ \boxed{y = \frac{17}{4} = 4,25}$