`lim_(xtoa) (log(x-a))/(log(e^(x)-e^(a)))`

Explanation

To solve the limit

\[ \lim_{x \to a} \frac{\log(x-a)}{\log(e^x - e^a)} \]

we first recognize that as \( x \) approaches \( a \), both the numerator and denominator approach \( \log(0) \), which is undefined. This gives us an indeterminate form of \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the denominator separately.

### Step-by-Step Solution:

1. **Identify the limit**:
\[ \lim_{x \to a} \frac{\log(x-a)}{\log(e^x - e^a)} \]

2. **Check the form**:
As \( x \to a \), both \( \log(x-a) \) and \( \log(e^x - e^a) \) approach \( \log(0) \), confirming we have the \( \frac{0}{0} \) form.

3. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
- The derivative of the numerator \( \log(x-a) \) is:
\[ \frac{d}{dx} \log(x-a) = \frac{1}{x-a} \]
- The derivative of the denominator \( \log(e^x - e^a) \) is:
\[ \frac{d}{dx} \log(e^x - e^a) = \frac{e^x}{e^x - e^a} \]

4. **Rewrite the limit**:
Now we can rewrite the limit as:
\[ \lim_{x \to a} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} = \lim_{x \to a} \frac{e^x - e^a}{(x-a)e^x} \]

5. **Evaluate the new limit**:
As \( x \to a \), both the numerator and denominator again approach \( 0 \), so we apply L'Hôpital's Rule again:
- The derivative of the numerator \( e^x - e^a \) is:
\[ e^x \]
- The derivative of the denominator \( (x-a)e^x \) using the product rule is:
\[ e^x + (x-a)e^x = e^x(1 + (x-a)) \]

6. **Rewrite the limit again**:
Now we have:
\[ \lim_{x \to a} \frac{e^x}{e^x(1 + (x-a))} = \lim_{x \to a} \frac{1}{1 + (x-a)} \]

7. **Evaluate the limit**:
As \( x \to a \):
\[ \frac{1}{1 + (a-a)} = \frac{1}{1} = 1 \]

### Final Answer:
Thus, the limit is:
\[ \lim_{x \to a} \frac{\log(x-a)}{\log(e^x - e^a)} = 1 \]