Given that `log_(a^(2))(a^(2)+1)=16` find the value of `log_(a^(32))"("a+1/a")"`

Answer

To solve the problem, we need to find the value of \( \log_{a^{32}} \left( a + \frac{1}{a} \right) \) given that \( \log_{a^2}(a^2 + 1) = 16 \).

### Step-by-step Solution:

1. **Understanding the Given Information:**
We know:
\[ \log_{a^2}(a^2 + 1) = 16 \]
This means:
\[ a^2 + 1 = (a^2)^{16} = a^{32} \]

2. **Rearranging the Expression:**
We want to find:
\[ \log_{a^{32}} \left( a + \frac{1}{a} \right) \]
Let's denote this value as \( x \):
\[ x = \log_{a^{32}} \left( a + \frac{1}{a} \right) \]

3. **Using Logarithmic Properties:**
We can use the property of logarithms:
\[ \log_{b^m}(a) = \frac{1}{m} \log_b(a) \]
Thus, we can rewrite \( x \):
\[ x = \frac{1}{32} \log_{a} \left( a + \frac{1}{a} \right) \]

4. **Simplifying \( a + \frac{1}{a} \):**
We can express \( a + \frac{1}{a} \) in terms of \( a^2 \):
\[ a + \frac{1}{a} = \frac{a^2 + 1}{a} \]

5. **Applying Logarithmic Properties Again:**
Using the property \( \log(a/b) = \log(a) - \log(b) \):
\[ \log_{a} \left( a + \frac{1}{a} \right) = \log_{a} \left( \frac{a^2 + 1}{a} \right) = \log_{a}(a^2 + 1) - \log_{a}(a) \]
Since \( \log_{a}(a) = 1 \):
\[ \log_{a} \left( a + \frac{1}{a} \right) = \log_{a}(a^2 + 1) - 1 \]

6. **Substituting the Known Value:**
From the given information, we know:
\[ \log_{a}(a^2 + 1) = 16 \cdot \log_{a}(a^2) = 16 \cdot 2 = 32 \]
Therefore:
\[ \log_{a}(a^2 + 1) = 32 \]
So:
\[ \log_{a} \left( a + \frac{1}{a} \right) = 32 - 1 = 31 \]

7. **Final Calculation:**
Now substituting back into our expression for \( x \):
\[ x = \frac{1}{32} \cdot 31 = \frac{31}{32} \]

### Final Answer:
\[ \log_{a^{32}} \left( a + \frac{1}{a} \right) = \frac{31}{32} \]