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Ayudemen porfavor . Determine 4 ala potencia x + 4 ala potencia -x, si se sabe que 2ala potencia x + 2 ala potencia -x =3

Sagot :

facundo3141592

Answer:

Es igual a 7.

Step-by-step explanation:

Ok, sabemos que:

[tex]2^x + 2^{-x} = 3[/tex]

Queremos calcular:

[tex]4^x + 4^{-x}[/tex]

Tambien sabemos que 4 = 2*2

Entonces:

[tex]4^x + 4^{-x} = (2*2)^x + (2*2)^{-x}[/tex]

y sabemos que 2*2 = 2^2

Entonces:

[tex]4^x + 4^{-x} = (2*2)^x + (2*2)^{-x} = (2^2)^x + (2^2)^{-x}[/tex]

y ahora podemos usar la relación:

[tex](a^n)^m = (a^m)^n = a^{m*n}[/tex]

entonces:

[tex](2^2)^x + (2^2)^{-x} = (2^x)^2 + (2^{-x})^2[/tex]

Ahora podemos completar cuadrados sumando y restando el termino:

[tex]2*(2^x)*(2^{-x})[/tex]

Asi tendremos:

[tex](2^x)^2 + (2^{-x})^2 = (2^x)^2 + (2^{-x})^2 + 2*(2^x)*(2^{-x}) - 2*(2^x)*(2^{-x}) = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})[/tex]

Entonces de momento tenemos que:

[tex]4^x + 4^{-x} = (2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x})[/tex]

Sabemos que el termino que esta dentro del parentesis es igual a 3.

Y tambien podemos usar la propiedad:

[tex]a^n*a^m = a^{n + m}[/tex]

en el termino de la derecha.

Asi tendremos:

[tex](2^x + 2^{-x})^2 - 2*(2^x)*(2^{-x}) = (3)^2 - 2*2^{x + (-x)} = 9 - 2 = 7[/tex]

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