Integral value of `x` which satisfies the equation `=log_6 54+(log)_x 16=(log)_(sqrt(2))x-(log)_(36)(4/9)i s ddot`

Explanation

To solve the equation \( \log_6 54 + \log_x 16 = \log_{\sqrt{2}} x - \log_{36} \left(\frac{4}{9}\right) \), we will follow these steps:

### Step 1: Simplify the Left Side
We can rewrite the left side of the equation:
\[ \log_6 54 + \log_x 16 = \log_6 (54) + \frac{\log 16}{\log x} \]
Using the property of logarithms, \( \log_a b + \log_a c = \log_a (bc) \), we can express \( \log_6 54 \) as:
\[ \log_6 (6 \cdot 9) = \log_6 6 + \log_6 9 = 1 + \log_6 9 \]
Thus, the left side becomes:
\[ 1 + \log_6 9 + \frac{4 \log 2}{\log x} \]

### Step 2: Simplify the Right Side
Now, we simplify the right side:
\[ \log_{\sqrt{2}} x - \log_{36} \left(\frac{4}{9}\right) \]
We can rewrite \( \log_{\sqrt{2}} x \) using the change of base formula:
\[ \log_{\sqrt{2}} x = \frac{\log x}{\log \sqrt{2}} = \frac{\log x}{\frac{1}{2} \log 2} = \frac{2 \log x}{\log 2} \]
Next, we simplify \( \log_{36} \left(\frac{4}{9}\right) \):
\[ \log_{36} \left(\frac{4}{9}\right) = \log_{36} \left(\frac{2^2}{3^2}\right) = \log_{36} (2^2) - \log_{36} (3^2) = \frac{2 \log 2}{\log 36} - \frac{2 \log 3}{\log 36} = \frac{2 (\log 2 - \log 3)}{\log 36} \]

### Step 3: Set the Equation
Now we have:
\[ 1 + \log_6 9 + \frac{4 \log 2}{\log x} = \frac{2 \log x}{\log 2} - \frac{2 (\log 2 - \log 3)}{\log 36} \]

### Step 4: Substitute and Solve for \( x \)
Let \( t = \log x \). The equation becomes:
\[ 1 + \log_6 9 + \frac{4 \log 2}{t} = \frac{2t}{\log 2} - \frac{2 (\log 2 - \log 3)}{\log 36} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{4 \log 2}{t} + \frac{2 (\log 2 - \log 3)}{\log 36} + 1 + \log_6 9 = \frac{2t}{\log 2} \]

### Step 6: Solve the Quadratic Equation
After simplifying, we can express this as a quadratic equation in terms of \( t \):
\[ 2t^2 - 2t - 4 = 0 \]
Dividing by 2:
\[ t^2 - t - 2 = 0 \]
Factoring gives:
\[ (t - 2)(t + 1) = 0 \]
Thus, \( t = 2 \) or \( t = -1 \).

### Step 7: Find \( x \)
1. If \( t = 2 \):
\[ \log x = 2 \implies x = 2^2 = 4 \]
2. If \( t = -1 \):
\[ \log x = -1 \implies x = 2^{-1} = \frac{1}{2} \]

### Final Answer
The integral value of \( x \) that satisfies the equation is:
\[ \boxed{4} \]