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

To solve the equation \(8^{2x} - 16 \cdot 8^x + 48 = 0\), we will use a substitution method. Let's assume \(y = 8^x\). This substitution transforms the equation into a quadratic form:

• \((8^x)^2 - 16 \cdot 8^x + 48 = 0\)
• \(\Rightarrow y^2 - 16y + 48 = 0\)

Now, this is a standard quadratic equation of the form \(ay^2 + by + c = 0\), where \(a = 1\), \(b = -16\), and \(c = 48\). We can use the quadratic formula to solve for \(y\):

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting the values of \(a\), \(b\), and \(c\):

• \(y = \frac{16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1}\)
• \(y = \frac{16 \pm \sqrt{256 - 192}}{2}\)
• \(y = \frac{16 \pm \sqrt{64}}{2}\)
• \(y = \frac{16 \pm 8}{2}\)

This gives us two solutions for \(y\):

• \(y = \frac{16 + 8}{2} = 12\)
• \(y = \frac{16 - 8}{2} = 4\)

Reverting the substitution \(y = 8^x\), we have two equations:

• \(8^x = 12\)
• \(8^x = 4\)

Solving for \(x\) in each equation:

• \(x = \log_8(12)\)
• \(x = \log_8(4)\)

To find the sum of all solutions, we calculate:

• \(\log_8(12) + \log_8(4)\)
• Using the property \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\), we have: \(\log_8(48)\)

Finally, since \(48 = 6 \times 8\), we can express this in base 6:

• \(\log_8(48) = \log_8(6 \times 8) = \log_8(6) + \log_8(8)\)
• Since \(\log_8(8) = 1\), we have \(\log_8(48) = 1 + \log_6(6)\)

Thus, the sum of all solutions is \(1 + \log_6(6)\), confirming that the correct answer is:

$1 + \log_6(6)$

Answer 2

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: \(1 + \log_8(6)\).