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 employ a substitution. Let \(y = 8^x\). This transforms the equation into the quadratic form:

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

This is a standard quadratic equation \(ay^2 + by + c = 0\), with \(a = 1\), \(b = -16\), and \(c = 48\). We apply the quadratic formula to find \(y\):

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

Substituting the coefficients:

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

This yields two solutions for \(y\):

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

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

• \(8^x = 12\)
• \(8^x = 4\)

Solving for \(x\) in each case yields:

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

The sum of all solutions is calculated as:

• \(\log_8(12) + \log_8(4)\)
• Applying the logarithm property \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\), we get: \(\log_8(48)\)

Since \(48 = 6 \times 8\), we can rewrite this as:

• \(\log_8(48) = \log_8(6 \times 8) = \log_8(6) + \log_8(8)\)
• As \(\log_8(8) = 1\), the expression becomes \(\log_8(48) = 1 + \log_8(6)\)

Therefore, the sum of all solutions is \(1 + \log_8(6)\).

$1 + \log_8(6)$