Let's carefully analyze the properties of the function [tex]\( f(x) = \ln(x) \)[/tex] and its inverse [tex]\( f^{-1}(x) = e^x \)[/tex].
### Step-by-Step Analysis:
1. Understanding the Function [tex]\( f(x) = \ln(x) \)[/tex]:
- Domain: The natural logarithm function, [tex]\( \ln(x) \)[/tex], is defined only for [tex]\( x > 0 \)[/tex]. Thus, the domain of [tex]\( f(x) = \ln(x) \)[/tex] is all real numbers greater than 0: [tex]\((0, \infty)\)[/tex].
- Range: The output of [tex]\( \ln(x) \)[/tex] can be any real number, as it spans from [tex]\(-\infty\)[/tex] to [tex]\(+\infty\)[/tex] as [tex]\( x \)[/tex] moves from very close to 0 to positive infinity. Therefore, the range of [tex]\( f(x) = \ln(x) \)[/tex] is all real numbers: [tex]\((-\infty, \infty)\)[/tex].
2. Finding the Inverse Function [tex]\( f^{-1}(x) = e^x \)[/tex]:
- When finding the inverse, the roles of the domain and range are swapped.
3. Understanding the Function [tex]\( f^{-1}(x) = e^x \)[/tex]:
- Domain: Since the range of the original function [tex]\( f(x) \)[/tex] was all real numbers [tex]\((-\infty, \infty)\)[/tex], the domain of [tex]\( f^{-1}(x) = e^x \)[/tex] will be all real numbers: [tex]\((-\infty, \infty)\)[/tex].
- Range: Since the domain of the original function [tex]\( f(x) \)[/tex] was all real numbers greater than zero [tex]\((0, \infty)\)[/tex], the range of [tex]\( f^{-1}(x) = e^x \)[/tex] will be all real numbers greater than 0: [tex]\((0, \infty)\)[/tex].
### Conclusion:
From the above analysis, we can conclude:
- The domain of [tex]\( f^{-1}(x) = e^x \)[/tex] is all real numbers [tex]\((-\infty, \infty)\)[/tex].
- The range of [tex]\( f^{-1}(x) = e^x \)[/tex] is all real numbers greater than 0 [tex]\((0, \infty)\)[/tex].
Hence, the correct conclusion is:
The domain of [tex]\( f^{-1}(x) \)[/tex] is all real numbers and the range is all real numbers greater than 0.
Thus, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
