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si: \( \quad b^{a}=5 \quad \wedge \quad a^{-b}=\frac{1}{3} \) Calcular: \( \quad I=b^{a^{b+1}} \)

Ask by Clarke Wood. in Spain
Jan 19,2025

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\( I = 125 \)

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Para resolver las ecuaciones \( b^a = 5 \) y \( a^{-b} = \frac{1}{3} \), primero despejemos \( a \) en la segunda ecuación: \[ a^{-b} = \frac{1}{3} \implies a = 3^{-1/b} \implies a = \frac{1}{3^{1/b}}. \] Sustituyamos \( a \) en la primera ecuación: \[ b^{\frac{1}{3^{1/b}}} = 5. \] Tomemos logaritmos en ambos lados para facilitar el cálculo: \[ \frac{1}{3^{1/b}} \cdot \log(b) = \log(5) \implies \log(b) = 3^{1/b} \cdot \log(5). \] Esto puede ser algo complejo de resolver directamente. Sin embargo, ahora busquemos la expresión para \( I \): \[ I = b^{a^{b+1}}. \] Primero, calculemos \( a^{b + 1} \): \[ a^{b+1} = a^b \cdot a = (b^{-1} + \log(3)) \cdot a, \] donde para encontrar un valor más específico de \( I \), podemos utilizar \( b^a = 5 \) dado que \( a = 3^{-1/b} \). Ahora, tomemos el logaritmo de \( I \): \[ \log(I) = a^{b + 1} \cdot \log(b), \] donde sabemos que \( b = 5^{1/a} \) por la relación inicial \( b^a=5 \). Por lo tanto, substituyendo obtenemos: \[ I = 5^{(3^{-1/b})^{b + 1}}. \] Finalmente, podemos resumir que: \[ I = 5^{\frac{2}{3}}. \]

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