Кто умеет решать такие , помогите пожалуйста)

Question 1

Question 2

В 1 системе 10 в какой степени ? Не очень видно.

Question 3

1 + lg(x+y)

Question 4

хотя бы как начать?

Question 5

Решите задачу:

$1)\quad \left \{ {{10^{1+lg(x+y)}=50} \atop {lg(x+y)+lg(x-y)=2-lg5}} \right. \; \; ODZ:\; \left \{ {{x\ \textgreater \ -y} \atop {x\ \textgreater \ y}} \right. \\\\ \left \{ {{10\cdot 10^{lg(x+y)}=50} \atop {lg[(x+y)(x-y)]=lg10^2-lg5}} \right. \; ,\; \; a^{log_{a}b}=b\; \to \; 10^{lgb}=b\\\\\left \{ {{10^{lg(x+y)}=50:10} \atop {lg[(x+y)(x-y)]=lg\frac{100}{5}}} \right. \; \left \{ {{x+y=5} \atop {(x+y)(x-y)=20}} \right. \; \left \{ {{x+y=5} \atop {5(x-y)=20}} \right. \; \left \{ {{x+y=5} \atop {x-y=4}} \right.$

$\left \{ {{2x=9} \atop {2y=1}} \right. \; \left \{ {{x=4,5} \atop {y=0,5}} \right. \; \; \; Otvet:\; \; (4,5\, ;\, 0,5)\, .\\\\2)\quad \left \{ {{lgx-lgy=lg15-1} \atop {10^{lg(3x+2y)}=39}} \right. \; ,\ \; \; ODZ:\; \; \left \{ {{x\ \textgreater \ 0,\; y\ \textgreater \ 0} \atop {x\ \textgreater \ -\frac{2y}{3}}} \right. \\\\ \left \{ {{lg\frac{x}{y}=lg15-lg10} \atop {3x+2y=39}} \right. \; \left \{ {lg\frac{x}{y}=lg\frac{15}{10}} \atop {3x+2y=39}} \right. \; \left \{ {{\frac{x}{y}=\frac{3}{2}} \atop {3x+2y=39}} \right.$

$\left \{ {{x=\frac{3y}{2}} \atop {3\cdot \frac{3y}{2}+2y=39}} \right. \; \left \{ {{x=\frac{3y}{2}} \atop {\frac{13y}{2}=39}} \right. \; \left \{ {{x=\frac{3\cdot 6}{2}=9} \atop {y=\frac{2\cdot 39}{13}=6}} \right. \\\\Otvet:\; (9,6)\; .$