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Ответ:

Решаем с помощью замены переменных .

[tex]\left\{\begin{array}{l}\bf 2^{x}+3^{y}=8\dfrac{1}{9}\\\bf 2^{x}\cdot 3^{y}-\dfrac{8}{9} \end{array}\right\ \ \ \ \bf p=2^{x}\ ,\ q=3^{y}\ \ \ \ \left\{\begin{array}{l}\bf p+q=\dfrac{73}{9}\\\bf p\cdot q=\dfrac{8}{9}\end{array}\right[/tex]  

[tex]\left\{\begin{array}{l}\bf p+\dfrac{8}{9p}=\dfrac{73}{9}\\\bf \ \ \ q=\dfrac{8}{9p}\end{array}\right\ \ \left\{\begin{array}{l}\bf \dfrac{9p^2+8}{9p}=\dfrac{73}{9}\\\bf \ \ \ q=\dfrac{8}{9p}\end{array}\right\ \ \left\{\begin{array}{l}\bf 9p^2+8=73p\\\bf \ \ \ q=\dfrac{8}{9p}\end{array}\right[/tex]  

[tex]\left\{\begin{array}{l}\bf 9\, p^2-73p+8=0\\\bf \ \ \ q=\dfrac{8}{9p}\end{array}\right\ \ \ \bf 9\, p^2-73\, p+8=0\ ,\\\\\\D=b^2-4ac=73^2-4\cdot 9\cdot 8=5041=71^2\ \ ,\\\\p_1=\dfrac{73-71}{18}=\dfrac{1}{9}\ \ ,\ \ p_2=\dfrac{73+71}{18}=8\\\\\\q_1=\dfrac{8}{9p_1}=8\ \ ,\ \ \ q_2=\dfrac{8}{9p_2}=\dfrac{1}{9}[/tex]      

Вернёмся к старым переменным .

[tex]\bf a)\ \ p_1=2^{x}=\dfrac{1}{9}\ \ ,\ \ q_1=3^{y}=8\ \ \ \Rightarrow \\\\x=log_2\dfrac{1}{9}=-2\, log_23\ \ \ ,\ \ \ y=log_3\, 8=3log_3\, 2\\\\\\b)\ \ p_2=2^{x}=8\ \ ,\ \ \ q_2=3^{y}=\dfrac{1}{9}\ \ \ \ \Rightarrow \\\\x=log_2\, 8=3\, log_2\. 2=3\ \ \ ,\ \ y=log_3\, \dfrac{1}{9}=-2\, log_3\, 3=-2\\\\\\Otvet:\ (-2\, log_23\ ;\ 3\, log_3\, 2\ )\ ,\ (\ 3\ ;-2\ )\ .[/tex]