The function `f(x)=log(x+sqrt(x^(2)+1))`, is
(a) an even function
(b) an odd function
(c ) a periodic function
(d) Neither an even nor an odd function.

Answer

To determine the nature of the function \( f(x) = \log(x + \sqrt{x^2 + 1}) \), we need to check whether it is an even function, an odd function, a periodic function, or neither.

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \log(x + \sqrt{x^2 + 1}) \]

2. **Calculate \( f(-x) \)**:
\[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}) \]

3. **Simplify \( f(-x) \)**:
We can rewrite \( f(-x) \) using properties of logarithms:
\[ f(-x) = \log(-x + \sqrt{x^2 + 1}) = \log\left(\sqrt{x^2 + 1} - x\right) \]

4. **Express \( f(-x) \) in terms of \( f(x) \)**:
We know that:
\[ f(x) = \log(x + \sqrt{x^2 + 1}) \]
Now, we can use the property of logarithms:
\[ f(-x) = \log\left(\sqrt{x^2 + 1} - x\right) = \log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) = -\log(x + \sqrt{x^2 + 1}) = -f(x) \]

5. **Check the condition for odd function**:
Since we have shown that:
\[ f(-x) = -f(x) \]
This satisfies the condition for \( f(x) \) to be an odd function.

6. **Conclusion**:
Therefore, the function \( f(x) \) is an odd function.

### Final Answer:
(b) an odd function