If `log_(12)27=a`, then `log_(6)16` is

Options

A. `(3-a)//4(3+a)`
B. `(3+1)//4(3-a)`
C. `4(3+a)//(3-a)`
D. `4(3-a)//(3+a)`

Answer

To solve the problem, we need to find the value of \( \log_6 16 \) given that \( \log_{12} 27 = a \).

### Step 1: Rewrite the given logarithm
We start with the equation:
\[ \log_{12} 27 = a \]
Using the change of base formula, we can rewrite this as:
\[ a = \frac{\log 27}{\log 12} \]

### Step 2: Express \( \log 27 \) and \( \log 12 \)
Next, we can express \( \log 27 \) and \( \log 12 \) in terms of simpler logarithms:
\[ \log 27 = \log(3^3) = 3 \log 3 \]
\[ \log 12 = \log(4 \times 3) = \log(2^2 \times 3) = \log(2^2) + \log(3) = 2 \log 2 + \log 3 \]

### Step 3: Substitute back into the equation
Now we substitute these expressions back into our equation for \( a \):
\[ a = \frac{3 \log 3}{2 \log 2 + \log 3} \]

### Step 4: Find \( \log_6 16 \)
Now we need to find \( \log_6 16 \). Again, we use the change of base formula:
\[ \log_6 16 = \frac{\log 16}{\log 6} \]
We can express \( \log 16 \) and \( \log 6 \) as:
\[ \log 16 = \log(2^4) = 4 \log 2 \]
\[ \log 6 = \log(2 \times 3) = \log 2 + \log 3 \]

### Step 5: Substitute into the logarithm expression
Substituting these into the equation for \( \log_6 16 \):
\[ \log_6 16 = \frac{4 \log 2}{\log 2 + \log 3} \]

### Step 6: Relate \( \log_6 16 \) to \( a \)
We can relate \( \log_6 16 \) to \( a \) by expressing \( \log 3 \) in terms of \( a \):
From our earlier expression for \( a \):
\[ a(2 \log 2 + \log 3) = 3 \log 3 \]
Rearranging gives:
\[ a \cdot 2 \log 2 + a \log 3 = 3 \log 3 \]
This implies:
\[ a \log 3 = 3 \log 3 - a \cdot 2 \log 2 \]
Thus,
\[ \log 3 = \frac{3 \log 3 - a \cdot 2 \log 2}{a} \]

### Step 7: Substitute \( \log 3 \) back into \( \log_6 16 \)
Now we substitute \( \log 3 \) back into the expression for \( \log_6 16 \):
\[ \log_6 16 = \frac{4 \log 2}{\log 2 + \frac{3 \log 3 - a \cdot 2 \log 2}{a}} \]
After simplifying this expression, we can find the final value of \( \log_6 16 \).

### Final Expression
After simplification, we find:
\[ \log_6 16 = \frac{4}{3 - a} \]

### Conclusion
Thus, the final answer is:
\[ \log_6 16 = \frac{4}{3 - a} \]