What is the value of `2 log_(8) 2-(1)/(3) log_(3) 9? `

Options

A. 0
B. 1
C. 2
D. `1//3`

Answer

To solve the expression \( 2 \log_{8} 2 - \frac{1}{3} \log_{3} 9 \), we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2 and base 3
We can use the change of base formula for logarithms:
\[ \log_{b} a = \frac{\log_{k} a}{\log_{k} b} \]
In our case, we can express \( \log_{8} 2 \) in terms of base 2:
\[ \log_{8} 2 = \frac{\log_{2} 2}{\log_{2} 8} \]
Since \( 8 = 2^3 \), we have:
\[ \log_{2} 8 = 3 \]
Thus:
\[ \log_{8} 2 = \frac{\log_{2} 2}{3} = \frac{1}{3} \]

### Step 2: Substitute back into the expression
Now substituting this back into the original expression:
\[ 2 \log_{8} 2 = 2 \cdot \frac{1}{3} = \frac{2}{3} \]

### Step 3: Simplify the second term
Next, we simplify \( \log_{3} 9 \):
\[ \log_{3} 9 = \log_{3} (3^2) = 2 \log_{3} 3 = 2 \]
Thus:
\[ -\frac{1}{3} \log_{3} 9 = -\frac{1}{3} \cdot 2 = -\frac{2}{3} \]

### Step 4: Combine the results
Now we combine both parts:
\[ \frac{2}{3} - \frac{2}{3} = 0 \]

### Final Answer
The value of \( 2 \log_{8} 2 - \frac{1}{3} \log_{3} 9 \) is \( 0 \).