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h) \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \)

Ask by Allan Mcfarlane. in Colombia
Feb 25,2025

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Answer

La respuesta simplificada es \( 6 a^{3.35} b^{2.25} \sqrt[5]{5} \).

Solution

Para resolver la expresión \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \), seguiremos los siguientes pasos: 1. **Descomponer cada raíz**: - Para \( \sqrt[5]{160 a^{8}} \): - Descomponemos \( 160 \) en sus factores primos: \( 160 = 2^5 \cdot 5 \). - Entonces, \( \sqrt[5]{160} = \sqrt[5]{2^5 \cdot 5} = 2 \cdot \sqrt[5]{5} \). - Para \( a^{8} \), aplicamos la propiedad de las raíces: \( \sqrt[5]{a^{8}} = a^{\frac{8}{5}} = a^{1.6} = a^{1} \cdot a^{0.6} \). - Para \( \sqrt[4]{81 a^{7} b^{9}} \): - Descomponemos \( 81 \) en sus factores primos: \( 81 = 3^4 \). - Entonces, \( \sqrt[4]{81} = 3 \). - Para \( a^{7} \), aplicamos la propiedad de las raíces: \( \sqrt[4]{a^{7}} = a^{\frac{7}{4}} = a^{1.75} = a^{1} \cdot a^{0.75} \). - Para \( b^{9} \), aplicamos la propiedad de las raíces: \( \sqrt[4]{b^{9}} = b^{\frac{9}{4}} = b^{2.25} = b^{2} \cdot b^{0.25} \). 2. **Combinar los resultados**: - Ahora combinamos los resultados de ambas raíces: \[ \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} = (2 \cdot \sqrt[5]{5} \cdot a^{1} \cdot a^{0.6}) \cdot (3 \cdot a^{1} \cdot a^{0.75} \cdot b^{2} \cdot b^{0.25}) \] 3. **Multiplicar los términos**: - Multiplicamos los coeficientes: \( 2 \cdot 3 = 6 \). - Multiplicamos las potencias de \( a \): \[ a^{1} \cdot a^{1} \cdot a^{0.6} \cdot a^{0.75} = a^{2 + 0.6 + 0.75} = a^{3.35} \] - Multiplicamos las potencias de \( b \): \[ b^{2} \cdot b^{0.25} = b^{2.25} \] 4. **Resultado final**: - La expresión completa se convierte en: \[ 6 \cdot a^{3.35} \cdot b^{2.25} \cdot \sqrt[5]{5} \] Por lo tanto, la solución final es: \[ 6 a^{3.35} b^{2.25} \sqrt[5]{5} \]

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Beyond the Answer

To simplify \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \), we'll start by breaking down the components. First, reduce each radical term: 1) For \( \sqrt[5]{160 a^{8}} \): - Break down \( 160 = 32 \times 5 = 2^5 \times 5 \). Thus, \( \sqrt[5]{160} = \sqrt[5]{2^5 \cdot 5} = 2 \sqrt[5]{5} \). - For \( a^{8} \), we have \( \sqrt[5]{a^{8}} = a^{8/5} = a^{1.6} = a^{1}a^{0.6} = a \sqrt[5]{a^3} \). Therefore, \( \sqrt[5]{160 a^{8}} = 2 a \sqrt[5]{5 a^{3}} \). 2) For \( \sqrt[4]{81 a^{7} b^{9}} \): - Break down \( 81 = 3^4 \). Thus, \( \sqrt[4]{81} = 3 \). - For \( a^{7} \), we have \( \sqrt[4]{a^{7}} = a^{7/4} = a^{1.75} = a^{1}a^{0.75} = a \sqrt[4]{a^3} \). - For \( b^{9} \), we have \( \sqrt[4]{b^{9}} = b^{9/4} = b^{2}b^{0.25} = b^{2} \sqrt[4]{b} \). Therefore, \( \sqrt[4]{81 a^{7} b^{9}} = 3 a b^{2} \sqrt[4]{a^3 b} \). Now, substituting back in gives us: \[ \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} = (2 a \sqrt[5]{5 a^{3}})(3 a b^{2} \sqrt[4]{a^3 b}) \] Next, multiplying the coefficients and like terms: \[ = 6 a^{2} b^{2} \sqrt[5]{5 a^{3}} \sqrt[4]{a^3 b} \] This is a good simplification step! This can be further reduced numerically by multiplying the radicals if required, but the result above captures the essence of the calculation. So, the final answer is: \[ 6a^{2}b^{2} \sqrt[5]{5 a^{3}} \sqrt[4]{a^3 b} \]

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