Question

The sum of all the solutions of the equation \[(8)^{2x} - 16 \cdot (8)^x + 48 = 0\]is:

• A. $1 + \log_6(8)$
• B. $\log_6(6)$
• C. $1 + \log_6(6)$
• D. $\log_6(4)$

Answer 1

The Correct Option is C

The given equation is:
\[ (8)^{2x} - 16 \cdot (8)^x + 48 = 0. \]
Substitute \(t = (8)^x\), which simplifies the equation to:
\[ t^2 - 16t + 48 = 0. \]
Solve the quadratic equation:
\[ t^2 - 16t + 48 = 0 \implies (t - 4)(t - 12) = 0. \]
Hence:
\[ t = 4 \quad \text{or} \quad t = 12. \]
Back-substituting \(t = (8)^x\), we get:
\[ (8)^x = 4 \implies x = \log_8(4), \]
\[ (8)^x = 12 \implies x = \log_8(12). \]
The sum of the solutions is:
\[ \text{Sum} = \log_8(4) + \log_8(12). \]
Using the logarithmic property \(\log_a(m) + \log_a(n) = \log_a(m \cdot n)\):
\[ \text{Sum} = \log_8(4 \cdot 12) = \log_8(48). \]
Express \(48\) as \(8 \cdot 6\):
\[ \log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6). \]
Since \(\log_8(8) = 1\), we have:
\[ \log_8(48) = 1 + \log_8(6). \]
Final Answer: \(1 + \log_8(6)\).