Question

La solución del sistema de ecuaciones exponenciales: 3^x-1 - 2^y=11. 3^x+ 2^y+1 =41

Answer 1

La solución del sistema de ecuaciones exponenciales planteada es:

y = $\frac{ln(8/5) }{ln(2)}$

x = $\frac{ln(11+2^{\frac{ln(8/5) }{ln(2)}})}{ln(3)}+1$

Dado,

$3^{x-1}-2^{y}=11$ (1)

$3^{x}+2^{y+1}=41$ (2)

Despejamos x de 1;

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

sumar $2^{y}$ a ambos lados;

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

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

Por propiedad de logaritmo natural;

Si f(x) = g(x), entonces ln(f(x)) = ln (g(x))

$ln(3^{x-1})=ln(11+2^{y})$

Aplicando la propiedad:$log_{a}(x^{b})  = b.log_{a}(x)$

$ln(3^{x-1})=ln(11+2^{y})$

$(x-1)ln(3)=ln(11+2^{y})$

Dividimos ambos lados entre ln(3);

$\frac{(x-1)ln(3)}{ln(3)} = \frac{ln(11+2^{y})}{ln(3)}$

x-1 = $\frac{ln(11+2^{y})}{ln(3)}$

Despejamos x;

x = $\frac{ln(11+2^{y})}{ln(3)}+1$

Sustituimos x en 2;

$3^{\frac{ln(11+2^{y})}{ln(3)}+1}+2^{y+1}=41$

Restamos $2^{y+1}$ a ambos lados;

$3^{\frac{ln(11+2^{y})}{ln(3)}+1}+2^{y+1}-2^{y+1}=41-2^{y+1}$

$3^{\frac{ln(11+2^{y})}{ln(3)}+1}=41-2^{y+1}$

Aplicando la propiedad:$log_{a}(x^{b})  = b.log_{a}(x)$

$ln(3^{\frac{ln(11+2^{y})}{ln(3)}+1})=ln(41-2^{y+1})$

Aplicando la propiedad:$log_{a}(x^{b})  = b.log_{a}(x)$

$(\frac{ln(11+2^{y})}{ln(3)}+1})ln(3)=ln(41-2^{y+1})$    

Desarrollar: $\frac{ln(11+2^{y})}{ln(3)}+1}$

= $(\frac{ln(11+2^{y}+ln(3))}{ln(3)})ln(3)=ln(41-2^{y+1})$

= $ln(11+2^{y})+ln(3)=ln(41-2^{y+1})$

Aplicamos propiedades de logaritmos: $log_{c}(a)+]log_{c}(b)  = log_{c}(ab)$

= $ln((11+2^{y}).3)=ln(41-2^{y+1})$

= $(11+2^{y}).3=41-2^{y+1}$

Sumamos  $2^{y+1}$ a ambos lados;

= $(11+2^{y}).3+2^{y+1}=41-2^{y+1}+2^{y+1}$

=  $(11+2^{y}).3+2^{y+1}=41$

= $33+3.2^{y}+2^{y+1}=41$

= $33+(3+2)2^{y}=41$

= $33+5.2^{y}=41$

= $5.2^{y}=41-33$

= $5.2^{y}=8$

= $2^{y}=8/5$

= $ln(2^{y})=ln(8/5)$

$yln(2)=ln(8/5)$

• y = $\frac{ln(8/5) }{ln(2)}$

Sustituyo y en x;

• x = $\frac{ln(11+2^{\frac{ln(8/5) }{ln(2)}})}{ln(3)}+1$