Solve `log_(6) 9-log_(9) 27 + log_(8)x = log_(64) x - log_(6) 4`..

Explanation

To solve the equation \( \log_{6} 9 - \log_{9} 27 + \log_{8} x = \log_{64} x - \log_{6} 4 \), we will use properties of logarithms step by step.

### Step 1: Rewrite the logarithms using the change of base formula
Using the property \( \log_{a} b = \frac{\log b}{\log a} \), we can rewrite the logarithms:
\[ \log_{6} 9 = \frac{\log 9}{\log 6}, \quad \log_{9} 27 = \frac{\log 27}{\log 9}, \quad \log_{8} x = \frac{\log x}{\log 8}, \quad \log_{64} x = \frac{\log x}{\log 64}, \quad \log_{6} 4 = \frac{\log 4}{\log 6} \]
Substituting these into the equation gives:
\[ \frac{\log 9}{\log 6} - \frac{\log 27}{\log 9} + \frac{\log x}{\log 8} = \frac{\log x}{\log 64} - \frac{\log 4}{\log 6} \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation to group the terms involving \( \log x \):
\[ \frac{\log 9}{\log 6} + \frac{\log 4}{\log 6} + \frac{\log x}{\log 8} - \frac{\log x}{\log 64} = \frac{\log 27}{\log 9} \]
This simplifies to:
\[ \frac{\log 9 + \log 4}{\log 6} + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{\log 27}{\log 9} \]

### Step 3: Combine the logarithms
Using the property \( \log a + \log b = \log(ab) \):
\[ \frac{\log(9 \times 4)}{\log 6} + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{\log 27}{\log 9} \]
Calculating \( 9 \times 4 = 36 \):
\[ \frac{\log 36}{\log 6} + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{\log 27}{\log 9} \]

### Step 4: Simplifying \( \log 27 \) and \( \log 36 \)
Now, we know \( \log 36 = \log(6^2) = 2 \log 6 \) and \( \log 27 = \log(3^3) = 3 \log 3 \):
\[ \frac{2 \log 6}{\log 6} + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{3 \log 3}{\log 9} \]
This simplifies to:
\[ 2 + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{3 \log 3}{2 \log 3} \]
Since \( \log 9 = 2 \log 3 \), we have:
\[ 2 + \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{3}{2} \]

### Step 5: Isolate \( \log x \)
Rearranging gives:
\[ \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = \frac{3}{2} - 2 \]
This simplifies to:
\[ \left(\frac{1}{\log 8} - \frac{1}{\log 64}\right) \log x = -\frac{1}{2} \]

### Step 6: Solve for \( \log x \)
Now, we can solve for \( \log x \):
\[ \log x = -\frac{1}{2} \cdot \frac{1}{\left(\frac{1}{\log 8} - \frac{1}{\log 64}\right)} \]

### Step 7: Calculate \( x \)
Finally, we can exponentiate to find \( x \):
\[ x = 8^{-1} = \frac{1}{8} \]

Thus, the solution is:
\[ \boxed{\frac{1}{8}} \]