The sum of all the solutions of the equation [(8)^2 x-16 (8)^x+48=0] is :

Question

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

Accepted Answer

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

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