Question:

<p><strong>Step 1: Remove the logarithm.</strong> Since both sides have logarithms, equate the arguments:</p><p>$(4^x - 2)^2 = 8(4^x - \frac{31}{8})$</p><p><strong>Step 2: Expand and simplify.</strong></p><p>$(4^x)^2 - 4·4^x + 4 = 8·4^x - 31$</p><p>$(4^x)^2 - 4·4^x - 8·4^x + 4 + 31 = 0$</p><p>$(4^x)^2 - 12·4^x + 35 = 0$</p><p><strong>Step 3: Use substitution $y = 4^x$.</strong></p><p>$y^2 - 12y + 35 = 0$</p><p><strong>Step 4: Solve the quadratic.</strong></p><p>$y = \frac{12 \pm \sqrt{144 - 140}}{2} = \frac{12 \pm 2}{2}$</p><p>$y_1 = 7, \quad y_2 = 5$</p><p><strong>Step 5: Verify validity.</strong> Both $4^x = 7 > 2$ and $4^x = 5 > 2$ satisfy the domain requirement (argument of logarithm must be positive).</p><p><strong>Step 6: Convert back to $x$ values.</strong></p><p>$4^{x_1} = 7 \implies x_1 = \log_4 7$</p><p>$4^{x_2} = 5 \implies x_2 = \log_4 5$</p><p><strong>Step 7: Find the sum of roots.</strong></p><p>$x_1 + x_2 = \log_4 7 + \log_4 5 = \log_4(7 \times 5) = \log_4 35$</p><p><strong>Step 8: Determine the interval.</strong> Since $4^1 = 4$ and $4^2 = 16$:</p><p>$4 < 35 < 16$ does not hold. Actually: $16 < 35 < 64 = 4^3$</p><p>So $2 < \log_4 35 < 3$, meaning $a = 2, b = 3$</p><p><strong>Step 9: Calculate final answer.</strong></p><p>$a + b = 2 + 3 = 5$</p><p><strong>∴ Answer:</strong> 5</p>