https://file.aiquickdraw.com/mathbot/file/172541460098647r8oqsi.jpg

Let's analyze the given problem step by step.

### Problem:
You are given:
- \( h(x) = e^{-x} \)
- \( k(x) = \ln x \)

You need to find \( h(k(x)) \).

### Solution:
To solve for \( h(k(x)) \), substitute \( k(x) \) into \( h(x) \):

\[
h(k(x)) = h(\ln x)
\]

Since \( h(x) = e^{-x} \), replace \( x \) in \( h(x) \) with \( \ln x \):

\[
h(k(x)) = e^{-\ln x}
\]

Now, apply the property of logarithms and exponentials:

\[
e^{-\ln x} = \frac{1}{e^{\ln x}} = \frac{1}{x}
\]

So, the answer is:

\[
h(k(x)) = \frac{1}{x}
\]

### Final Answer:
The correct choice is **A** \( \frac{1}{x} \).

Would you like more details or have any further questions?

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Here are 5 related questions you might find interesting:
1. How do the properties of logarithms simplify the expression \( e^{\ln x} \)?
2. What is the domain of the function \( h(k(x)) = \frac{1}{x} \)?
3. How would the composition change if \( k(x) = \log_b x \) instead of \( \ln x \)?
4. Can you find the inverse function of \( h(x) = e^{-x} \)?
5. What is the derivative of \( h(k(x)) \) with respect to \( x \)?

**Tip:** When composing functions, it's crucial to carefully substitute and apply the correct mathematical properties for simplification.

Given that h(x) = e^(-x) and k(x) = ln(x), x > 0. Find h(k(x)).

Learn how to solve the function composition problem h(k(x)) where h(x) = e^(-x) and k(x) = ln(x). This problem explores key concepts in exponential and logarithmic functions and is suitable for students in Grades 11-12.

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