If `("log"_(a)x)/("log"_(ab)x) = 4 + k + "log"_(a)b, "then" k=`

Options

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

Explanation

To solve the equation \(\frac{\log_a x}{\log_{ab} x} = 4 + k + \log_a b\), we will follow these steps:

### Step 1: Rewrite the logarithm
We can use the change of base formula for logarithms. The change of base formula states that:
\[ \log_{ab} x = \frac{\log_a x}{\log_a (ab)} \]
Using this, we can express \(\log_{ab} x\) as:
\[ \log_{ab} x = \frac{\log_a x}{\log_a a + \log_a b} = \frac{\log_a x}{1 + \log_a b} \]

### Step 2: Substitute into the equation
Now, substituting \(\log_{ab} x\) into our original equation gives us:
\[ \frac{\log_a x}{\frac{\log_a x}{1 + \log_a b}} = 4 + k + \log_a b \]
This simplifies to:
\[ \log_a x \cdot \frac{1 + \log_a b}{\log_a x} = 4 + k + \log_a b \]
Thus, we have:
\[ 1 + \log_a b = 4 + k + \log_a b \]

### Step 3: Simplify the equation
Next, we can subtract \(\log_a b\) from both sides:
\[ 1 = 4 + k \]

### Step 4: Solve for \(k\)
Now, we can isolate \(k\):
\[ k = 1 - 4 = -3 \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{-3} \]