Question
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Answer
\( n = \frac{\ln(26)}{\ln(4)} - 2 \)
Solution
1. Partimos de la ecuación:
\( 3 \cdot 4^{n+2}=78 \)
2. Dividimos ambos lados de la ecuación entre 3 para aislar la potencia:
\( 4^{n+2} = \frac{78}{3} = 26 \)
3. Aplicamos logaritmos en ambos lados de la ecuación para poder "bajar" el exponente. Utilizando logaritmo natural (aunque se puede usar cualquier base):
\( \ln(4^{n+2}) = \ln(26) \)
4. Utilizamos la propiedad de los logaritmos que establece que \( \ln(a^b) = b \ln(a) \):
\( (n+2) \ln(4) = \ln(26) \)
5. Despejamos \( n+2 \):
\( n+2 = \frac{\ln(26)}{\ln(4)} \)
6. Finalmente, despejamos \( n \):
\( n = \frac{\ln(26)}{\ln(4)} - 2 \)
Este es el valor de \( n \) en términos de logaritmos.
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Simplify this solution Beyond the Answer
To solve the equation \( 3 \cdot 4^{n+2} = 78 \), you can start by isolating the term with \( n \). Divide both sides by 3: \[ 4^{n+2} = \frac{78}{3} = 26 \] Next, take the logarithm of both sides to bring down the exponent: \[ \log(4^{n+2}) = \log(26) \] Using the property of logarithms that allows you to bring the exponent down, this becomes: \[ (n+2) \cdot \log(4) = \log(26) \] Now, isolate \( n+2 \): \[ n+2 = \frac{\log(26)}{\log(4)} \] Finally, solve for \( n \): \[ n = \frac{\log(26)}{\log(4)} - 2 \] Calculating the values using a calculator: \[ n \approx \frac{1.41497}{0.60206} - 2 \approx 2.35 - 2 \approx 0.35 \] So, \( n \approx 0.35 \) is the solution!
