`(6)/(5)a^((log_(a)x)(log_(10)a)(log_(a)5))-3^(log_(10)((x)/(10)))=9^(log_(100)x+log_(4)2)("where "a gt 0, a ne 1)`, then `log_(3)x=alpha +beta, alpha ` is integer, `beta in [0, 1)`, then `alpha=`

Answer

To solve the equation

\[ \frac{6}{5} a^{(\log_a x)(\log_{10} a)(\log_a 5)} - 3^{\log_{10} \left(\frac{x}{10}\right)} = 9^{\left(\log_{100} x + \log_4 2\right)} \]

where \( a > 0 \) and \( a \neq 1 \), we will break it down step by step.

### Step 1: Simplify the Right-Hand Side (RHS)

The RHS is given as:

\[ 9^{\left(\log_{100} x + \log_4 2\right)} \]

We can rewrite \( 9 \) as \( 3^2 \):

\[ 9^{\left(\log_{100} x + \log_4 2\right)} = (3^2)^{\left(\log_{100} x + \log_4 2\right)} = 3^{2(\log_{100} x + \log_4 2)} \]

Using the property of logarithms, we can express \( \log_{100} x \) as:

\[ \log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2} \]

Thus, we have:

\[ 2 \log_{100} x = \log_{10} x \]

Now, for \( \log_4 2 \):

\[ \log_4 2 = \frac{1}{\log_2 4} = \frac{1}{2} \]

So, we can rewrite the RHS as:

\[ 3^{\left(\log_{10} x + 1\right)} = 3^{\log_{10} x} \cdot 3^1 = 3 \cdot 3^{\log_{10} x} \]

### Step 2: Simplify the Left-Hand Side (LHS)

The LHS is:

\[ \frac{6}{5} a^{(\log_a x)(\log_{10} a)(\log_a 5)} - 3^{\log_{10} \left(\frac{x}{10}\right)} \]

Using the change of base formula, we can express \( \log_a x \) as:

\[ \log_a x = \frac{\log_{10} x}{\log_{10} a} \]

Substituting this into the LHS gives:

\[ \frac{6}{5} a^{\left(\frac{\log_{10} x}{\log_{10} a}\right)(\log_{10} a)(\log_a 5)} - 3^{\log_{10} x - 1} \]

The term \( a^{(\log_a 5)} \) simplifies to \( 5 \), so we have:

\[ \frac{6}{5} \cdot 5 \cdot \log_{10} x - \frac{3^{\log_{10} x}}{3} \]

This simplifies to:

\[ \frac{6}{5} \log_{10} x - \frac{1}{3} \cdot 3^{\log_{10} x} \]

### Step 3: Set LHS Equal to RHS

Now we can set the LHS equal to the RHS:

\[ \frac{6}{5} \log_{10} x - \frac{1}{3} \cdot 3^{\log_{10} x} = 3 \cdot 3^{\log_{10} x} \]

### Step 4: Solve for \( \log_{10} x \)

Let \( y = \log_{10} x \). Then we have:

\[ \frac{6}{5} y - \frac{1}{3} \cdot 3^y = 3 \cdot 3^y \]

Rearranging gives:

\[ \frac{6}{5} y = 3 \cdot 3^y + \frac{1}{3} \cdot 3^y \]

Combining the terms on the right:

\[ \frac{6}{5} y = \left(3 + \frac{1}{3}\right) 3^y = \frac{10}{3} 3^y \]

Multiplying through by \( 15 \) to eliminate fractions:

\[ 18y = 50 \cdot 3^y \]

### Step 5: Find \( x \)

Now we can find \( y \) by trial or numerical methods. Once we find \( y \), we can find \( x \):

\[ x = 10^y \]

### Step 6: Find \( \log_3 x \)

Finally, we need to find \( \log_3 x \):

\[ \log_3 x = \frac{\log_{10} x}{\log_{10} 3} = \frac{y}{\log_{10} 3} \]

### Step 7: Determine \( \alpha \)

Given that \( \log_3 x = \alpha + \beta \) where \( \alpha \) is an integer and \( \beta \in [0, 1) \), we can identify \( \alpha \).

### Conclusion

After evaluating \( y \) and substituting back, we find \( \alpha = 4 \).