The value of `log_((8-3sqrt7))(8+3sqrt7)` is

Options

A. -2
B. -1
C. 0
D. Not defined

Answer

To find the value of \( \log_{(8-3\sqrt{7})}(8+3\sqrt{7}) \), we can follow these steps:

### Step 1: Rationalize \( 8 + 3\sqrt{7} \)
We can express \( 8 + 3\sqrt{7} \) in terms of \( 8 - 3\sqrt{7} \) by multiplying by its conjugate:

\[ 8 + 3\sqrt{7} = \frac{(8 + 3\sqrt{7})(8 - 3\sqrt{7})}{8 - 3\sqrt{7}} = \frac{8^2 - (3\sqrt{7})^2}{8 - 3\sqrt{7}} \]

### Step 2: Calculate the numerator
Calculating the numerator:

\[ 8^2 = 64 \quad \text{and} \quad (3\sqrt{7})^2 = 9 \times 7 = 63 \]

So,

\[ 8^2 - (3\sqrt{7})^2 = 64 - 63 = 1 \]

### Step 3: Substitute back into the expression
Now substituting back, we have:

\[ 8 + 3\sqrt{7} = \frac{1}{8 - 3\sqrt{7}} \]

### Step 4: Rewrite the logarithm
Now we can rewrite the logarithm:

\[ \log_{(8-3\sqrt{7})}(8 + 3\sqrt{7}) = \log_{(8-3\sqrt{7})}\left(\frac{1}{8 - 3\sqrt{7}}\right) \]

### Step 5: Use logarithmic properties
Using the property of logarithms, \( \log_b\left(\frac{1}{a}\right) = -\log_b(a) \):

\[ \log_{(8-3\sqrt{7})}\left(\frac{1}{8 - 3\sqrt{7}}\right) = -\log_{(8-3\sqrt{7})}(8 - 3\sqrt{7}) \]

### Step 6: Evaluate the logarithm
Since \( \log_{(8-3\sqrt{7})}(8 - 3\sqrt{7}) = 1 \) (because any number to the base of itself is 1):

\[ -\log_{(8-3\sqrt{7})}(8 - 3\sqrt{7}) = -1 \]

### Final Answer
Thus, the value of \( \log_{(8-3\sqrt{7})}(8+3\sqrt{7}) \) is:

\[ \boxed{-1} \]