Lo que pretendo demostrar es que: en la resolución de un ejercicio del Álgebra, lo importante es saber como se llega a su resultado, para que haya un verdadero aprendizaje. Mi Email es jorgecarrillom@gmail.com

Ecuaciones Exponenciales.

Son las ecuaciones en que la incógnita es el exponente de una cantidad.

Para resolver este tipo de ecuaciones, se aplican logaritmos a los dos miembros de la ecuación y se despeja la incógnita.

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Procedimiento:

1°) Se aplica la fórmula para el logaritmo de una potencia.

2°) Se busca el logaritmo del otro miembro de la ecuación.

3°) Encontrado los logaritmos se procede a realizar operaciones.

4°) Se despejan la incógnita (el exponente de la potencia).

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Ejemplos:

 a) Resolver la ecuación  3ˣ = 60

> Aplicando logaritmos:

x(Log 3) = Log 60

x(0.477121) = 1.778151

x = 1.778151/0.477121

x = 3.72   Solución.

b) Resolver la ecuación 5²ˣ⁻¹ = 125

> Aplicando logaritmos:

2x-1(Log 5) = Log 125

2x-1(0.698970) = 2.096910

2x = (2.096919/0.698970) +1

x = 3+1 /2

x = 2   Solución.

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Ejercicio 301.

1) Resolver  5ˣ =3

> Aplicando logaritmos:

x(log 5) = log 3

x = log 3/log 5

x = 0.477121/0.698970

x = 0.6826  Solución.

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3) Resolver 0.2ˣ = 0.0016

> Aplicando logaritmos:

x (log 0.2) = log 0.0016

x = log 0.0016/log 0.2

x = -2.795988/-0.698970

x = 4.  Solución.

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5) Resolver 3ˣ⁺¹ = 729

> Aplicando logaritmos:

x+1(log 3) = log 729

x = (log 729/log 3)-1

x = (2.862727/0.477121)-1

x = 6-1 = 5  Solución.

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7) Resolver  2³ˣ⁺¹ = 128

> Aplicando logaritmos:

3x+1(log 2) = log 128

3x+1 = log 128/log 2

x = [(log 128/log 2)-1]/3

x = [(2.107210/0.301030)-1]/3

x = (7-1)/3

x = 6/3 =2  Solución.

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Comentarios en: "Ecuaciones Exponenciales." (3)

  1. Anónimo dijo:

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  2. Anónimo dijo:

    me la pela

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