Question

alguien me ayuda pls​

Answer 1

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${3}^{x + 2} + {2}^{y} = 11$

${3}^{x} + {2}^{y - 4} = \frac{5}{6}$

En la primera ecuacion:

${3}^{x + 2} + {2}^{y} = 11 \\ {3}^{x} \times {3}^{2} + {2}^{y} = 11$

En la segunda:

${3}^{x} + {2}^{y} \times {2}^{ - 4} = \frac{5}{6} \\ {3}^{x} \times {3}^{2} + {2}^{y} \times \frac{9}{16} = \frac{15}{2}$

Ahora restamos la primera ecuacion con la segunda:

${3}^{x} \times {3}^{2} - {3}^{x} \times {3}^{2} + {2}^{y} - 2 ^{y} \times \frac{9}{16} = 11 - \frac{15}{2} \\ {2}^{y} (1 - \frac{9}{16} ) = \frac{22 - 15}{2} \\ {2}^{y} ( \frac{7}{16} ) = \frac{7}{2} \\ {2}^{y} = 8 \\ log_{2}( {2}^{y} ) = log_{2}( {2}^{3} ) \\ y = 3$

Con el valor de Y podemos calcular de una forma mas simple el de X en la primera ecuacion:

${3}^{x + 2} + {2}^{y} = 11 \\ {3}^{x + 2} + {2}^{3} = 11 \\ {3}^{x + 2} + 8 = 11 \\ {3}^{x + 2} = 3 {}^{1} \\ x + 2 = 1 \\ x = - 1$