We are tasked with solving the equation:

$\frac{2 \log_3(x) - 4}{\log_3\left(\frac{x}{9}\right)} = \log_3\left(x^7\right) - \left(\frac{1}{\log_x(3)} - 8\right)$

Let's go step by step to simplify and solve this equation.

Step 1: Simplify the logarithmic terms

Using logarithmic properties, we can simplify the terms:

• $\log_3\left(\frac{x}{9}\right)$ can be rewritten as: $\log_3\left(\frac{x}{9}\right) = \log_3(x) - \log_3(9)$ Since $\log_3(9) = 2$, this becomes: $\log_3\left(\frac{x}{9}\right) = \log_3(x) - 2$

• $\log_3(x^7)$ can be rewritten as: $\log_3(x^7) = 7 \log_3(x)$

• $\log_x(3)$ can be rewritten using the change of base formula: $\log_x(3) = \frac{1}{\log_3(x)}$

Step 2: Substitute these into the original equation

We now have the simplified equation:

$\frac{2 \log_3(x) - 4}{\log_3(x) - 2} = 7 \log_3(x) - \left(\frac{1}{\frac{1}{\log_3(x)}} - 8\right)$

Simplifying the term $\frac{1}{\frac{1}{\log_3(x)}}$ results in just $\log_3(x)$. So the equation becomes:

$\frac{2 \log_3(x) - 4}{\log_3(x) - 2} = 7 \log_3(x) - (\log_3(x) - 8)$

Step 3: Simplify further

The right-hand side simplifies as:

$7 \log_3(x) - (\log_3(x) - 8) = 7 \log_3(x) - \log_3(x) + 8 = 6 \log_3(x) + 8$

Thus, the equation becomes:

$\frac{2 \log_3(x) - 4}{\log_3(x) - 2} = 6 \log_3(x) + 8$

Step 4: Solve the equation

We now multiply both sides by $\log_3(x) - 2$ to eliminate the denominator on the left-hand side:

$2 \log_3(x) - 4 = (6 \log_3(x) + 8)(\log_3(x) - 2)$

Expand the right-hand side:

$2 \log_3(x) - 4 = (6 \log_3(x)^2 - 12 \log_3(x) + 8 \log_3(x) - 16)$

Simplify the terms:

$2 \log_3(x) - 4 = 6 \log_3(x)^2 - 4 \log_3(x) - 16$

Move all terms to one side of the equation:

$0 = 6 \log_3(x)^2 - 6 \log_3(x) - 12$

Step 5: Solve the quadratic equation

We can divide the entire equation by 6 to simplify:

$0 = \log_3(x)^2 - \log_3(x) - 2$

This is a quadratic equation in terms of $\log_3(x)$. Using the quadratic formula:

$\log_3(x) = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$\log_3(x) = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}$

Thus, we have two solutions for $\log_3(x)$:

$\log_3(x) = \frac{1 + 3}{2} = 2 \quad \text{or} \quad \log_3(x) = \frac{1 - 3}{2} = -1$

Step 6: Find $x$

• If $\log_3(x) = 2$, then $x = 3^2 = 9$.
• If $\log_3(x) = -1$, then $x = 3^{-1} = \frac{1}{3}$.

Final Solution

The two possible solutions are:

$x = 9 \quad \text{or} \quad x = \frac{1}{3}$

Would you like further details or have any questions?

Related Questions:

1. What are the properties of logarithms used in solving equations like this?
2. How does the change of base formula for logarithms work?
3. Can you explain how to solve quadratic equations in more detail?
4. What are some common mistakes to avoid when solving logarithmic equations?
5. How does the quadratic formula apply to logarithmic functions?

Tip:

Always verify the solutions by substituting them back into the original equation to check for extraneous solutions.