If `3log_((3x^(2)))27-2log_((3x))9=0`, then what is the value of x?

Options

A. `1//243`
B. `1//7`
C. `1//49`
D. None of these

Answer

To solve the equation \( 3\log_{(3x^2)}27 - 2\log_{(3x)}9 = 0 \), we will follow these steps:

### Step 1: Rewrite the logarithms in terms of a common base
We can express the logarithms using the change of base formula. We know that:

\[ \log_{(3x^2)}27 = \frac{\log_{3}27}{\log_{3}(3x^2)} \]
and
\[ \log_{(3x)}9 = \frac{\log_{3}9}{\log_{3}(3x)} \]

### Step 2: Substitute the values of the logarithms
We know that \( \log_{3}27 = 3 \) (since \( 27 = 3^3 \)) and \( \log_{3}9 = 2 \) (since \( 9 = 3^2 \)). Thus, we can substitute these values:

\[ \log_{(3x^2)}27 = \frac{3}{\log_{3}(3x^2)} = \frac{3}{\log_{3}3 + \log_{3}x^2} = \frac{3}{1 + 2\log_{3}x} \]

And similarly for \( \log_{(3x)}9 \):

\[ \log_{(3x)}9 = \frac{2}{\log_{3}(3x)} = \frac{2}{\log_{3}3 + \log_{3}x} = \frac{2}{1 + \log_{3}x} \]

### Step 3: Substitute back into the original equation
Now we substitute these back into the original equation:

\[ 3 \cdot \frac{3}{1 + 2\log_{3}x} - 2 \cdot \frac{2}{1 + \log_{3}x} = 0 \]

This simplifies to:

\[ \frac{9}{1 + 2\log_{3}x} - \frac{4}{1 + \log_{3}x} = 0 \]

### Step 4: Find a common denominator and simplify
To combine the fractions, we find a common denominator:

\[ \frac{9(1 + \log_{3}x) - 4(1 + 2\log_{3}x)}{(1 + 2\log_{3}x)(1 + \log_{3}x)} = 0 \]

This leads to:

\[ 9 + 9\log_{3}x - 4 - 8\log_{3}x = 0 \]

### Step 5: Combine like terms
Combining like terms gives:

\[ 5 + \log_{3}x = 0 \]

### Step 6: Solve for \(\log_{3}x\)
Thus, we have:

\[ \log_{3}x = -5 \]

### Step 7: Convert back to exponential form
Now, converting back to exponential form gives:

\[ x = 3^{-5} = \frac{1}{3^5} = \frac{1}{243} \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{\frac{1}{243}} \]