Решите систему уравнений 3^log3(x+y)=4
2^x + 2^y = 10

$\left \{ {{3^{log3(x+y)}=4} \atop {2^x + 2^y = 10}} \right.$

ОДЗ: $x+y\ \textgreater \ 0$

$\left \{ {{x+y}=4} \atop {2^x + 2^y = 10}} \right.$

$\left \{ {{x}=4-y} \atop {2^{4-y} + 2^y = 10}} \right.$

$\left \{ {{x}=4-y} \atop {2^{4}*2^{-y} + 2^y = 10}} \right.$

$\left \{ {{x}=4-y} \atop {16* \frac{1}{2^y} + 2^y = 10}} \right.$

$16* \frac{1}{2^y} + 2^y = 10}$

Замена: 2^y=t, $t\ \textgreater \ 0$

$16* \frac{1}{t} + t = 10$

$\frac{16}{t} + t- 10=0$

t^2- 10t+16=0

D_1=5^2-1*16=25-16=9

t_1=5+3=8

t_2=5-3=2

$\left \{ {{x=4-y} \atop {2^y=8}} \right.$ или $\left \{ {{x=4-y} \atop {2^y=2}} \right.$

$\left \{ {{x=4-y} \atop {2^y=2^3}} \right.$ или $\left \{ {{x=4-y} \atop {2^y=2^1}} \right.$

$\left \{ {{y=3} \atop {x=1}} \right.$ или $\left \{ {{y=1} \atop {x=3}} \right.$

Ответ: (1;3); (3;1)