Question

${9}^{x} - {2}^{y} = 1 \\ {9}^{ - x} - 2 ^{ - y} = - \frac{1}{6}$

Answer 1

${9}^{x} - {2}^{y} = 1 \\ \frac{1}{ {9}^{x} } - \frac{1}{ {2}^{y} } = - \frac{1}{6}$


${2}^{y} = {9}^{x} - 1 \\ \frac{1}{ {9}^{x} } - \frac{1}{ {9}^{x} - 1} = - \frac{1}{6}$

$\frac{ {9}^{x} - 1 - {9}^{x} }{ {9}^{x}( {9}^{x} - 1) } = - \frac{1}{6}$

$\frac{ - 1}{ {9}^{x} ( {9}^{x} - 1) } = - \frac{1}{6}$
${9}^{x} ( {9}^{x} - 1) = 6$
пусть
${9}^{x} = t > 0$
${t}^{2} - t - 6 = 0$

$t_{1} = - 2 \\ t_{2} =3$

первый корень не удовлетворяет условию t>0.

${9}^{x} = 3 \\ ({ {3}^{2} })^{x} = {3}\\ {3}^{2x} = {3}^{1} \\ 2x = 1 \\ x = 0.5$
${2}^{y} = {9}^{0.5} - 1 \\ {2}^{y} = 3 - 1 \\ {2}^{y} = 2 \\ {2}^{y} = {2}^{1} \\ y = 1$
ответ: (0,5; 1)