The roots of the equation `2^(x+2).3^((3x)/(x-1))=9` are given by

Options

A. `log_(2)((2)/(3)),-2`
B. 3,-3
C. `-2,1-(log3)/(log 2)`
D. `1-log((2)/(3)),2`

Explanation

To solve the equation \( 2^{(x+2)} \cdot 3^{\frac{3x}{x-1}} = 9 \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting \( 9 \) as \( 3^2 \):
\[ 2^{(x+2)} \cdot 3^{\frac{3x}{x-1}} = 3^2 \]

### Step 2: Take logarithm on both sides
Taking logarithm on both sides, we have:
\[ \log(2^{(x+2)} \cdot 3^{\frac{3x}{x-1}}) = \log(3^2) \]

### Step 3: Apply logarithmic properties
Using the properties of logarithms, we can expand both sides:
\[ \log(2^{(x+2)}) + \log(3^{\frac{3x}{x-1}}) = 2 \log(3) \]
This simplifies to:
\[ (x+2) \log(2) + \frac{3x}{x-1} \log(3) = 2 \log(3) \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ (x+2) \log(2) + \frac{3x}{x-1} \log(3) - 2 \log(3) = 0 \]

### Step 5: Combine terms
We can factor out \( \log(3) \):
\[ (x+2) \log(2) + \log(3) \left( \frac{3x}{x-1} - 2 \right) = 0 \]

### Step 6: Set each factor to zero
This gives us two equations to solve:
1. \( x + 2 = 0 \)
2. \( \frac{3x}{x-1} - 2 = 0 \)

### Step 7: Solve the first equation
From the first equation:
\[ x + 2 = 0 \implies x = -2 \]

### Step 8: Solve the second equation
From the second equation:
\[ \frac{3x}{x-1} = 2 \]
Cross-multiplying gives:
\[ 3x = 2(x - 1) \implies 3x = 2x - 2 \implies x = -2 \]

Now substituting \( x = -2 \) back into the second equation:
\[ \frac{3(-2)}{-2 - 1} = 2 \implies \frac{-6}{-3} = 2 \implies 2 = 2 \text{ (True)} \]

### Step 9: Find the second root
Now we rearrange the second equation:
\[ \frac{3x - 2(x - 1)}{x - 1} = 0 \implies 3x - 2x + 2 = 0 \implies x + 2 = 0 \implies x = -2 \]

Now, we can also express \( x - 1 = -\frac{\log(3)}{\log(2)} \):
\[ x = 1 - \frac{\log(3)}{\log(2)} \]

### Conclusion
Thus, the roots of the equation are:
1. \( x = -2 \)
2. \( x = 1 - \frac{\log(3)}{\log(2)} \)