Let a and b be real numbers greater than 1 for which there exists a positive real number c, different from 1, such that `2(log_a c +log_b c)=9log_ab c`. Find the largest possible value of `log_a b`.

Explanation

To solve the problem, we need to analyze the given equation and manipulate it using properties of logarithms. Here’s a step-by-step solution:

### Step 1: Write down the given equation
We start with the equation provided in the problem:
\[ 2(\log_a c + \log_b c) = 9 \log_{ab} c \]

### Step 2: Use the change of base formula
Using the change of base formula, we can express the logarithms in terms of natural logarithms or common logarithms:
\[ \log_a c = \frac{\log c}{\log a} \]
\[ \log_b c = \frac{\log c}{\log b} \]
\[ \log_{ab} c = \frac{\log c}{\log(ab)} = \frac{\log c}{\log a + \log b} \]

Substituting these into the equation gives:
\[ 2\left(\frac{\log c}{\log a} + \frac{\log c}{\log b}\right) = 9 \cdot \frac{\log c}{\log a + \log b} \]

### Step 3: Factor out \(\log c\)
Assuming \(\log c \neq 0\) (since \(c \neq 1\)), we can divide both sides by \(\log c\):
\[ 2\left(\frac{1}{\log a} + \frac{1}{\log b}\right) = \frac{9}{\log a + \log b} \]

### Step 4: Simplify the left-hand side
The left-hand side can be combined:
\[ 2\left(\frac{\log b + \log a}{\log a \cdot \log b}\right) = \frac{9}{\log a + \log b} \]

This simplifies to:
\[ \frac{2(\log a + \log b)}{\log a \cdot \log b} = \frac{9}{\log a + \log b} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 2(\log a + \log b)^2 = 9 \log a \cdot \log b \]

### Step 6: Let \(x = \log_b a\)
Let \(x = \log_b a\), then \(\log a = \frac{\log b}{x}\). Substituting this into the equation gives:
\[ 2\left(\frac{\log b}{x} + \log b\right)^2 = 9 \cdot \frac{\log b}{x} \cdot \log b \]

### Step 7: Simplify further
Let \(y = \log b\):
\[ 2\left(\frac{y}{x} + y\right)^2 = 9 \cdot \frac{y^2}{x} \]
This expands to:
\[ 2\left(\frac{y(1+x)}{x}\right)^2 = 9 \cdot \frac{y^2}{x} \]

### Step 8: Cancel \(y^2\) and simplify
Assuming \(y \neq 0\):
\[ 2\left(\frac{(1+x)^2}{x^2}\right) = \frac{9}{x} \]

### Step 9: Rearranging
Multiplying through by \(x^2\) gives:
\[ 2(1 + 2x + x^2) = 9x \]
\[ 2 + 4x + 2x^2 = 9x \]
Rearranging gives:
\[ 2x^2 - 5x + 2 = 0 \]

### Step 10: Factor or use the quadratic formula
Factoring gives:
\[ (2x - 1)(x - 2) = 0 \]
Thus, \(x = \frac{1}{2}\) or \(x = 2\).

### Step 11: Conclusion
The largest possible value of \(\log_b a\) is:
\[ \boxed{2} \]