Solve the Logarithmic Equation: log₄(x²+8x+1)/log₄8 = 2

To solve the problem, we'll follow these steps:

• Step 1: Simplify the given expression using logarithmic identities.
• Step 2: Solve the resulting quadratic equation for $x$.

Now, let's work through each step:

Step 1: We start with the equation:

$\frac{\log_4(x^2+8x+1)}{\log_4 8} = 2$

We know that $\log_4 8 = \frac{3}{2}$, since $8 = 4^{3/2}$. Thus, we can rewrite the equation as:

$\log_4(x^2+8x+1) = 2 \times \frac{3}{2} = 3$

Applying the property of logarithms that states $\log_b a = c \Rightarrow a = b^c$, we have:

$x^2 + 8x + 1 = 4^3 = 64$

Step 2: Solve the resulting quadratic equation:

$x^2 + 8x + 1 = 64$

Subtract 64 from both sides to bring the equation to standard form:

$x^2 + 8x + 1 - 64 = 0$

$x^2 + 8x - 63 = 0$

Now, apply the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 8$, and $c = -63$:

$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1}$

$x = \frac{-8 \pm \sqrt{64 + 252}}{2}$

$x = \frac{-8 \pm \sqrt{316}}{2}$

Simplify $\sqrt{316}$ as $\sqrt{79 \cdot 4} = 2\sqrt{79}$:

$x = \frac{-8 \pm 2\sqrt{79}}{2}$

Thus, $x = -4 \pm \sqrt{79}$.

Therefore, the solution to the equation is $x = -4 \pm \sqrt{79}$.