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

Answer: (C)

$1 + \log_6(6)$

Answer 2

The given equation is:
$(8)^{2x} - 16 \cdot (8)^x + 48 = 0.$
Substitute $t = (8)^x$, which simplifies the equation to:
$t^2 - 16t + 48 = 0.$
Solve the quadratic equation:
$t^2 - 16t + 48 = 0 \implies (t - 4)(t - 12) = 0.$
Hence:
$t = 4 \quad \text{or} \quad t = 12.$
Back-substituting $t = (8)^x$, we get:
$(8)^x = 4 \implies x = \log_8(4),$
$(8)^x = 12 \implies x = \log_8(12).$
The sum of the solutions is:
$\text{Sum} = \log_8(4) + \log_8(12).$
Using the logarithmic property $\log_a(m) + \log_a(n) = \log_a(m \cdot n)$:
$\text{Sum} = \log_8(4 \cdot 12) = \log_8(48).$
Express $48$ as $8 \cdot 6$:
$\log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6).$
Since $\log_8(8) = 1$, we have:
$\log_8(48) = 1 + \log_8(6).$
Final Answer: $1 + \log_8(6)$.