If `"log"_(3) x + "log"_(9)x^(2) + "log"_(27)x^(3) = 9`, then x =

Options

A. 3
B. 9
C. 27
D. none of these

Explanation

To solve the equation \( \log_3 x + \log_9 x^2 + \log_{27} x^3 = 9 \), we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 3

We know that:
- \( \log_9 x^2 = \frac{\log_3 x^2}{\log_3 9} \)
- \( \log_{27} x^3 = \frac{\log_3 x^3}{\log_3 27} \)

Now, we can express \( \log_3 9 \) and \( \log_3 27 \):
- \( \log_3 9 = \log_3 (3^2) = 2 \)
- \( \log_3 27 = \log_3 (3^3) = 3 \)

Thus, we can rewrite the logarithms:
\[ \log_9 x^2 = \frac{2 \log_3 x}{2} = \log_3 x \]
\[ \log_{27} x^3 = \frac{3 \log_3 x}{3} = \log_3 x \]

### Step 2: Substitute back into the original equation

Now we substitute these back into the original equation:
\[ \log_3 x + \log_3 x + \log_3 x = 9 \]
This simplifies to:
\[ 3 \log_3 x = 9 \]

### Step 3: Solve for \( \log_3 x \)

Dividing both sides by 3:
\[ \log_3 x = 3 \]

### Step 4: Convert from logarithmic to exponential form

Using the property of logarithms that states if \( \log_b a = c \), then \( a = b^c \):
\[ x = 3^3 \]
Calculating this gives:
\[ x = 27 \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{27} \]