QJEEM24Ma5830: JEE Main 2024 Mathematics - Class 11 JEEM24Ma5830 Solution: The sum of all the solutions of the equation is :

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

$y^2 - 16y + 48 = 0$

Now, we have a quadratic equation in $y$ . To find the roots of this quadratic equation, we can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$ , $b = -16$ , and $c = 48$ . Substituting these values into the formula, we get:

$y = \frac{16 \pm \sqrt{256 - 192}}{2}$

$y = \frac{16 \pm \sqrt{64}}{2}$

$y = \frac{16 \pm 8}{2}$

Solving for the two possible values of $y$ , we have:

$y = \frac{16 + 8}{2} = 12$

$y = \frac{16 - 8}{2} = 4$

Now, recall that we substituted $y = (8)^x$ . So, we need to solve for $x$ when $y = 12$ and $y = 4$ :

$8^x = 12$

$x = \log_8(12)$

$8^x = 4$

$x = \log_8(4)$

Therefore, the solutions for $x$ are $\log_8(12)$ and $\log_8(4)$ . The sum of these solutions is:

$\log_8(12) + \log_8(4)$

Using the logarithmic property that $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ , we get:

$\log_8(12) + \log_8(4) = \log_8(12 \cdot 4)$

$= \log_8(48)$

Now, we note that:

$48 = 8 \cdot 6$

Thus,

$\log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6)$

Therefore, the sum of all the solutions of the equation is:

Option A $1 + \log_8(6)$ .