((2log3(x))-4)/log3(x/9)=7log3(x)-(1/logx(3))^2-8

We are given the equation:

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

### Step 1: Simplify both sides.

#### Left-hand side (LHS):

Using the property \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\):

\[
\log_3\left(\frac{x}{9}\right) = \log_3(x) - \log_3(9)
\]
Since \(\log_3(9) = 2\), we get:
\[
\log_3\left(\frac{x}{9}\right) = \log_3(x) - 2
\]
Thus, the LHS becomes:
\[
\frac{2 \log_3(x) - 4}{\log_3(x) - 2}
\]

#### Right-hand side (RHS):

Next, simplify the RHS. First, simplify \(\frac{1}{\log_x(3)}\). Using the change of base formula, we know that:
\[
\log_x(3) = \frac{1}{\log_3(x)}
\]
Thus:
\[
\frac{1}{\log_x(3)} = \log_3(x)
\]
Therefore:
\[
\left(\frac{1}{\log_x(3)}\right)^2 = \left(\log_3(x)\right)^2
\]
The RHS becomes:
\[
7 \log_3(x) - \left(\log_3(x)\right)^2 - 8
\]

### Step 2: Rewrite the equation.

Now, the equation is:

\[
\frac{2 \log_3(x) - 4}{\log_3(x) - 2} = 7 \log_3(x) - \left(\log_3(x)\right)^2 - 8
\]

### Step 3: Substitute \( y = \log_3(x) \).

Let \( y = \log_3(x) \), then the equation becomes:

\[
\frac{2y - 4}{y - 2} = 7y - y^2 - 8
\]

### Step 4: Solve the equation.

Multiply both sides by \( y - 2 \) (assuming \( y \neq 2 \)):

\[
2y - 4 = (7y - y^2 - 8)(y - 2)
\]

Expand the right-hand side:

\[
2y - 4 = (7y^2 - 14y - y^3 + 2y^2 - 8y + 16)
\]
\[
2y - 4 = -y^3 + 9y^2 - 22y + 16
\]

Bring all terms to one side:

\[
0 = -y^3 + 9y^2 - 24y + 20
\]
\[
y^3 - 9y^2 + 24y - 20 = 0
\]

### Step 5: Solve the cubic equation.

Now, we need to solve \( y^3 - 9y^2 + 24y - 20 = 0 \). We can attempt to factor this or use the Rational Root Theorem to find possible roots.

The possible rational roots are \( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \).

Testing \( y = 2 \):

\[
2^3 - 9(2^2) + 24(2) - 20 = 8 - 36 + 48 - 20 = 0
\]

Thus, \( y = 2 \) is a solution.

### Step 6: Factor the cubic polynomial.

Since \( y = 2 \) is a root, we can factor \( y^3 - 9y^2 + 24y - 20 \) as \( (y - 2)(y^2 - 7y + 10) \).

Now, solve \( y^2 - 7y + 10 = 0 \) using the quadratic formula:

\[
y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)}
\]
\[
y = \frac{7 \pm \sqrt{49 - 40}}{2}
\]
\[
y = \frac{7 \pm \sqrt{9}}{2}
\]
\[
y = \frac{7 \pm 3}{2}
\]

Thus, \( y = \frac{7 + 3}{2} = 5 \) and \( y = \frac{7 - 3}{2} = 2 \).

### Step 7: Find \( x \).

Recall that \( y = \log_3(x) \), so:

- If \( y = 2 \), then \( \log_3(x) = 2 \), which means \( x = 3^2 = 9 \).
- If \( y = 5 \), then \( \log_3(x) = 5 \), which means \( x = 3^5 = 243 \).

Thus, the solutions are \( x = 9 \) and \( x = 243 \).

### Final Answer:

The solutions are \( x = 9 \) and \( x = 243 \).

---

Would you like more details on any of these steps or help with related problems? Here are some related questions for further exploration:

1. How does the change of base formula work for logarithms?
2. What is the Rational Root Theorem, and how is it applied to solve cubic equations?
3. How do you factor cubic polynomials in general?
4. Can you solve logarithmic equations with different bases?
5. How can you use logarithms in real-world applications?

**Tip:** Always check if substituting your solutions back into the original equation satisfies all conditions.

This detailed solution tackles a complex logarithmic equation, involving the change of base formula, rational root theorem, and solving a cubic equation. Learn step-by-step methods to simplify logarithmic expressions and solve for x, with solutions x = 9 and x = 243. Ideal for students in Grades 10-12.

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