Question

Решить систему уравнений
3^x * 7^y = 63
3^x + 7^y = 16

Answer 1

$\left\{\begin{array}{l}3^{x}\cdot 7^{y}=63\\3^{x}+7^{y}=16\end{array}\right\ \ \left\{\begin{array}{l}t=3^{x}\\p=7^{y}\end{array}\right\ \ \left\{\begin{array}{l}t\cdot p=63\\t+p=16\end{array}\right\ \ \left\{\begin{array}{l}t\, (16-t)=63\\p=16-t\end{array}\right$

$\left\{\begin{array}{l}t^2-16t+63=0\\p=16-t\end{array}\right\ \ \left\{\begin{array}{l}t_1=7\ ,\ t_2=9\\p_1=9\ ,\ p_2=7\end{array}\right\\\\\\a)\ \ \left\{\begin{array}{l}3^{x}=7\\7^{y}=9\end{array}\right\ \ \left\{\begin{array}{l}x=log_37\\y=log_79\end{array}\right$

$b)\ \ \left\{\begin{array}{l}3^{x}=9\\7^{y}=7\end{array}\right\ \ \left\{\begin{array}{l}3^{x}=3^2\\7^{y}=7^1\end{array}\right\ \ \left\{\begin{array}{l}x=2\\y=1\end{array}\right\\\\\\Otvet:\ \ (\ log_37\ ;\ log_79\ )\ ,\ (\ 2\ ;\ 1\ )$