If `(log)_4 5=aa n d(log)_5 6=b ,` then `(log)_3 2` is equal to `1/(2a+1)` (b) `1/(2b+1)` (c) `2a b+1` (d) `1/(2a b-1)`

Options

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

Explanation

To solve the problem, we need to find the value of \( \log_3 2 \) given that \( \log_4 5 = a \) and \( \log_5 6 = b \).

### Step-by-step Solution:

1. **Express \( 5 \) in terms of \( 4 \)**:
From the equation \( \log_4 5 = a \), we can rewrite it using the definition of logarithms:
\[ 5 = 4^a \]

2. **Express \( 6 \) in terms of \( 5 \)**:
Similarly, from the equation \( \log_5 6 = b \):
\[ 6 = 5^b \]

3. **Substitute \( 5 \) into the equation for \( 6 \)**:
Now, substitute the expression for \( 5 \) from step 1 into the equation for \( 6 \):
\[ 6 = (4^a)^b = 4^{ab} \]

4. **Express \( 4 \) in terms of \( 2 \)**:
We know that \( 4 = 2^2 \), so we can rewrite \( 6 \):
\[ 6 = (2^2)^{ab} = 2^{2ab} \]

5. **Break down \( 6 \)**:
We can also express \( 6 \) as \( 2 \times 3 \):
\[ 6 = 2^1 \times 3^1 \]

6. **Equate the two expressions for \( 6 \)**:
Now we have two expressions for \( 6 \):
\[ 2^{2ab} = 2^1 \times 3^1 \]
From this, we can equate the powers of \( 2 \):
\[ 2ab = 1 + \log_2 3 \]

7. **Express \( \log_2 3 \)**:
Rearranging gives us:
\[ \log_2 3 = 2ab - 1 \]

8. **Express \( \log_3 2 \)**:
We need to find \( \log_3 2 \). Using the change of base formula:
\[ \log_3 2 = \frac{1}{\log_2 3} \]
Substituting the value of \( \log_2 3 \) from the previous step:
\[ \log_3 2 = \frac{1}{2ab - 1} \]

### Conclusion:
Thus, the value of \( \log_3 2 \) is:
\[ \log_3 2 = \frac{1}{2ab - 1} \]

### Final Answer:
The correct option is (d) \( \frac{1}{2ab - 1} \).