If `log_(3)30 =1/a` and `log_(5) 30 = 1/b` then the value of `3 log_(30)2` is:

Options

A. `3(1 + a +b)`
B. `2(1-a -b)`
C. `3(1-a-b)`
D. `3(1+a-b)`

Answer

To solve the problem, we need to find the value of \(3 \log_{30} 2\) given that \( \log_{3} 30 = \frac{1}{a} \) and \( \log_{5} 30 = \frac{1}{b} \).

### Step-by-Step Solution:

1. **Convert the given logarithms**:
We know that:
\[ \log_{3} 30 = \frac{1}{a} \implies \log_{30} 3 = a \]
\[ \log_{5} 30 = \frac{1}{b} \implies \log_{30} 5 = b \]

2. **Use the property of logarithms**:
We can express \(30\) in terms of its prime factors:
\[ 30 = 2 \times 3 \times 5 \]
Therefore, we can write:
\[ \log_{30} 30 = \log_{30} (2 \times 3 \times 5) \]

3. **Apply the logarithmic identity**:
Using the property of logarithms that states \(\log_{a}(b \cdot c) = \log_{a} b + \log_{a} c\), we have:
\[ \log_{30} 30 = \log_{30} 2 + \log_{30} 3 + \log_{30} 5 \]
Since \(\log_{30} 30 = 1\), we can write:
\[ 1 = \log_{30} 2 + a + b \]

4. **Rearranging the equation**:
From the equation above, we can express \(\log_{30} 2\):
\[ \log_{30} 2 = 1 - a - b \]

5. **Finding \(3 \log_{30} 2\)**:
Now we can find \(3 \log_{30} 2\):
\[ 3 \log_{30} 2 = 3(1 - a - b) = 3 - 3a - 3b \]

### Final Answer:
Thus, the value of \(3 \log_{30} 2\) is:
\[ 3 - 3a - 3b \]

---