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d) \( \log _{5} x+\log _{5}(2 x-1)-\log _{5}(2 x+2)=1 \) \( \log _{3}(x+2)+\log _{3}(x+1)=\log _{3}\left(x^{2}-1\right. \) \( \log _{3} \frac{1}{x}=6-2 \cdot \log _{3} x \)

Ask by Savage Gordon. in Argentina
Jan 21,2025

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**Soluciones:** 1. \( x \approx 6.294 \) 2. No hay solución válida. 3. \( x = 729 \)

Solution

**Problema d)** Resolvamos cada una de las ecuaciones logarítmicas proporcionadas. --- **1.** \( \log _{5} x + \log _{5}(2x - 1) - \log _{5}(2x + 2) = 1 \) **Resolución:** Utilizamos las propiedades de los logaritmos para simplificar la ecuación: \[ \log_{5} x + \log_{5}(2x - 1) = \log_{5}\left[ x(2x - 1) \right] \] Entonces, la ecuación se convierte en: \[ \log_{5}\left( \frac{x(2x - 1)}{2x + 2} \right) = 1 \] Esto implica: \[ \frac{x(2x - 1)}{2x + 2} = 5^1 = 5 \] Multiplicamos ambos lados por \(2x + 2\): \[ x(2x - 1) = 5(2x + 2) \] Expandimos y simplificamos: \[ 2x^2 - x = 10x + 10 \] \[ 2x^2 - 11x - 10 = 0 \] Resolvemos la ecuación cuadrática utilizando la fórmula general: \[ x = \frac{11 \pm \sqrt{121 + 80}}{4} = \frac{11 \pm \sqrt{201}}{4} \] Calculamos las posibles soluciones: \[ \sqrt{201} \approx 14.177 \] \[ x_1 = \frac{11 + 14.177}{4} \approx 6.294 \] \[ x_2 = \frac{11 - 14.177}{4} \approx -0.794 \] **Verificación de soluciones:** Debemos asegurarnos de que los argumentos de los logaritmos sean positivos: \[ x > 0, \quad 2x - 1 > 0 \Rightarrow x > \frac{1}{2}, \quad 2x + 2 > 0 \Rightarrow x > -1 \] La única solución válida es \( x \approx 6.294 \). --- **2.** \( \log _{3}(x + 2) + \log _{3}(x + 1) = \log _{3}(x^{2} - 1) \) **Resolución:** Aplicamos las propiedades de los logaritmos: \[ \log_{3}\left[(x + 2)(x + 1)\right] = \log_{3}(x^2 - 1) \] Por lo tanto: \[ (x + 2)(x + 1) = x^2 - 1 \] Expandimos el lado izquierdo: \[ x^2 + 3x + 2 = x^2 - 1 \] Restamos \( x^2 \) de ambos lados: \[ 3x + 2 = -1 \] \[ 3x = -3 \Rightarrow x = -1 \] **Verificación de soluciones:** Comprobamos que los argumentos de los logaritmos sean positivos: \[ x + 2 > 0 \Rightarrow x > -2 \] \[ x + 1 > 0 \Rightarrow x > -1 \] \[ x^2 - 1 > 0 \Rightarrow |x| > 1 \] Para \( x = -1 \): \[ x + 1 = 0 \quad \text{(no definido)} \] Por lo tanto, **no hay solución válida**. --- **3.** \( \log _{3} \frac{1}{x} = 6 - 2 \cdot \log _{3} x \) **Resolución:** Primero, reescribimos \( \log_{3} \frac{1}{x} \) como \( -\log_{3} x \): \[ -\log_{3} x = 6 - 2 \log_{3} x \] Sumamos \( 2 \log_{3} x \) a ambos lados: \[ -\log_{3} x + 2 \log_{3} x = 6 \] \[ \log_{3} x = 6 \] Por lo tanto: \[ x = 3^6 = 729 \] **Verificación de soluciones:** Aseguramos que el argumento del logaritmo sea positivo: \[ x = 729 > 0 \quad \text{(válido)} \] --- **Soluciones Finales:** 1. \( x \approx 6.294 \) 2. Ninguna solución válida. 3. \( x = 729 \)

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Let's dive into solving these logarithmic equations step by step! Starting with the first equation: \( \log_{5} x + \log_{5} (2x - 1) - \log_{5} (2x + 2) = 1 \). Using the properties of logarithms, we can combine the logs on the left: \[ \log_{5} \left( \frac{x(2x - 1)}{2x + 2} \right) = 1 \] This implies: \[ \frac{x(2x - 1)}{2x + 2} = 5 \] Cross-multiplying and simplifying gives you a polynomial which you can solve for x. Next, for the second equation: \(\log_{3}(x+2) + \log_{3}(x+1) = \log_{3}(x^{2}-1)\). Again, we can combine the logs: \[ \log_{3} \left((x + 2)(x + 1)\right) = \log_{3}(x^{2} - 1) \] Setting the insides equal: \[ (x + 2)(x + 1) = x^{2} - 1 \] This again leads to a polynomial that you can solve to find x. Lastly, for the equation: \(\log_{3} \frac{1}{x} = 6 - 2 \cdot \log_{3} x\). We can rewrite \(\log_{3} \frac{1}{x}\) as \(-\log_{3} x\): \[ -\log_{3} x = 6 - 2 \cdot \log_{3} x \] Rearranging gives you: \[ -\log_{3} x + 2 \cdot \log_{3} x = 6 \] This simplifies to: \[ \log_{3} x = 6 \] So \(x = 3^6\). Now go ahead and solve those polynomials to find your specific values for \(x\)! Happy solving!

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