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

<p>First, let's start by substituting $y = (8)^x$ in the given equation. By substituting, the equation $8^{2x} - 16 \cdot 8^x + 48 = 0$ will be transformed into</p> <p>$y^2 - 16y + 48 = 0$</p> <p>Now, we have a quadratic equation in $y$. To find the roots of this quadratic equation, we can use the quadratic formula:</p> <p>$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$</p> <p>In this case, $a = 1$, $b = -16$, and $c = 48$. Substituting these values into the formula, we get:</p> <p>$y = \frac{16 \pm \sqrt{256 - 192}}{2}$</p> <p>$y = \frac{16 \pm \sqrt{64}}{2}$</p> <p>$y = \frac{16 \pm 8}{2}$</p> <p>Solving for the two possible values of $y$, we have:</p> <p>$y = \frac{16 + 8}{2} = 12$</p> <p>$y = \frac{16 - 8}{2} = 4$</p> <p>Now, recall that we substituted $y = (8)^x$. So, we need to solve for $x$ when $y = 12$ and $y = 4$:</p> <p>$8^x = 12$</p> <p>$x = \log_8(12)$</p> <p>$8^x = 4$</p> <p>$x = \log_8(4)$</p> <p>Therefore, the solutions for $x$ are $\log_8(12)$ and $\log_8(4)$. The sum of these solutions is:</p> <p>$\log_8(12) + \log_8(4)$</p> <p>Using the logarithmic property that $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$, we get:</p> <p>$\log_8(12) + \log_8(4) = \log_8(12 \cdot 4)$</p> <p>$= \log_8(48)$</p> <p>Now, we note that:</p> <p>$48 = 8 \cdot 6$</p> <p>Thus,</p> <p>$\log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6)$</p> <p>Therefore, the sum of all the solutions of the equation is:</p> <p>Option A $1 + \log_8(6)$.</p>