If `log_(12) ` m = a and `log_(18)` m = b , then `(a- 2b)/(b - 2a)` is

Options

A. `log_(2)3`
B. 1
C. `log_(3)2`
D. none of these

Answer

To solve the problem, we start with the given equations:

1. \( \log_{12} m = a \)
2. \( \log_{18} m = b \)

We need to find the value of \( \frac{a - 2b}{b - 2a} \).

### Step 1: Express \( a \) and \( b \) in terms of natural logarithms
Using the change of base formula for logarithms, we can express \( a \) and \( b \) as follows:

\[ a = \frac{\log m}{\log 12} \]
\[ b = \frac{\log m}{\log 18} \]

### Step 2: Substitute \( a \) and \( b \) into the expression
Now, we substitute \( a \) and \( b \) into the expression \( \frac{a - 2b}{b - 2a} \):

\[ \frac{a - 2b}{b - 2a} = \frac{\frac{\log m}{\log 12} - 2 \cdot \frac{\log m}{\log 18}}{\frac{\log m}{\log 18} - 2 \cdot \frac{\log m}{\log 12}} \]

### Step 3: Factor out \( \log m \)
Since \( \log m \) is common in both the numerator and the denominator, we can factor it out:

\[ = \frac{\log m \left( \frac{1}{\log 12} - 2 \cdot \frac{1}{\log 18} \right)}{\log m \left( \frac{1}{\log 18} - 2 \cdot \frac{1}{\log 12} \right)} \]

Canceling \( \log m \) (assuming \( m > 0 \) and \( m \neq 1 \)) gives us:

\[ = \frac{\frac{1}{\log 12} - \frac{2}{\log 18}}{\frac{1}{\log 18} - \frac{2}{\log 12}} \]

### Step 4: Simplifying the expression
Now we can simplify the numerator and denominator:

Numerator:
\[ = \frac{\log 18 - 2 \log 12}{\log 12 \cdot \log 18} \]

Denominator:
\[ = \frac{\log 12 - 2 \log 18}{\log 12 \cdot \log 18} \]

Thus, we can rewrite the expression as:

\[ = \frac{\log 18 - 2 \log 12}{\log 12 - 2 \log 18} \]

### Step 5: Further simplification
Now we can express \( \log 18 \) and \( \log 12 \) in terms of their prime factors:

\[ \log 18 = \log(2 \cdot 3^2) = \log 2 + 2 \log 3 \]
\[ \log 12 = \log(2^2 \cdot 3) = 2 \log 2 + \log 3 \]

Substituting these into the expression gives:

Numerator:
\[ = \log 2 + 2 \log 3 - 2(2 \log 2 + \log 3) = \log 2 + 2 \log 3 - 4 \log 2 - 2 \log 3 = -3 \log 2 \]

Denominator:
\[ = 2 \log 2 + \log 3 - 2(\log 2 + 2 \log 3) = 2 \log 2 + \log 3 - 2 \log 2 - 4 \log 3 = -3 \log 3 \]

### Step 6: Final expression
Now we have:

\[ = \frac{-3 \log 2}{-3 \log 3} = \frac{\log 2}{\log 3} \]

### Final Answer
Thus, the value of \( \frac{a - 2b}{b - 2a} \) is:

\[ \frac{\log 2}{\log 3} \]