Question
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Si: \( m=\log _{5} 3 \) y \( n=\log _{5} 2 \) halle \( \log _{3} 72 \)

Ask by Ortega Valdez. in Peru
Mar 17,2025

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Answer

\[ \log_3 72 = \frac{3n}{m} + 2 \]

Solution

Sea \( 72 = 2^3 \cdot 3^2 \). Entonces, usando propiedades de los logaritmos, tenemos \[ \log_3 72 = \log_3 \left(2^3 \cdot 3^2\right) = \log_3 \left(2^3\right) + \log_3 \left(3^2\right). \] Utilizamos la propiedad del logaritmo del exponente: \[ \log_3 \left(2^3\right) = 3 \log_3 2 \quad \text{y} \quad \log_3 \left(3^2\right) = 2 \log_3 3. \] Como \(\log_3 3 = 1\), se tiene: \[ \log_3 72 = 3 \log_3 2 + 2. \] Ahora, usamos cambio de base para expresar \(\log_3 2\) en función de \( n = \log_5 2 \) y \( m = \log_5 3 \): \[ \log_3 2 = \frac{\log_5 2}{\log_5 3} = \frac{n}{m}. \] Substituyendo: \[ \log_3 72 = 3 \cdot \frac{n}{m} + 2 = \frac{3n}{m} + 2. \]

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The Deep Dive

Para calcular \( \log_{3} 72 \), primero descomponemos 72 en sus factores primos: \( 72 = 2^3 \times 3^2 \). Utilizando la propiedad de los logaritmos que dice que \( \log_{a}(b \cdot c) = \log_{a}(b) + \log_{a}(c) \), podemos escribir: \[ \log_{3} 72 = \log_{3}(2^3) + \log_{3}(3^2) = 3 \log_{3}(2) + 2 \log_{3}(3) = 3 \log_{3}(2) + 2. \] Ahora, sabemos que \( \log_{3}(2) \) se puede expresar usando el cambio de base en términos de \( m \) y \( n \): \[ \log_{3}(2) = \frac{\log_{5}(2)}{\log_{5}(3)} = \frac{n}{m}. \] Sustituyendo en la expresión anterior, obtenemos: \[ \log_{3} 72 = 3 \left(\frac{n}{m}\right) + 2. \] Finalmente, esta es la forma que nos permite calcular \( \log_{3} 72 \) en función de \( m \) y \( n \).

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