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The power rule is a helpful tool when dealing with logarithmic expressions that contain exponentiation. It allows us to simplify these expressions by pulling the exponent outside of the logarithm.
For a general logarithm, the power rule states:

• \( \ln(a^b) = b \cdot \ln(a) \)

This means that if you have a logarithm of a number raised to a power, you can remove the exponent from the inner expression and multiply the logarithm by it.
In the initial exercise, we applied the power rule to \( \ln(\sqrt[3]{e}) \), rewritten as \( \ln(e^{1/3}) \). By doing so, we took the exponent \( \frac{1}{3} \) out of the natural logarithm equation, resulting in \( \frac{1}{3} \cdot \ln(e) \). This greatly simplifies the expression before further computation.

Inverse functions are pairs of functions that "undo" each other's work. When it comes to logarithmic and exponential functions, they have a special relationship.
The natural logarithm \( \ln(x) \) and the exponential function \( e^x \) are inverse functions. This means they cancel each other out; hence:

• \( e^{\ln(x)} = x \)
• \( \ln(e^x) = x \)

This inverse property plays a crucial role in evaluating expressions involving these functions.
In the solution, the property \( \ln(e) = 1 \) is used. That's because applying the natural logarithm to the number \( e \) cancels out the exponential nature of \( e \), leaving us with the simplest form of \( e^1 = 1 \). Recognizing this relationship can help solve and simplify similar logarithmic and exponential problems with ease.

Simplifying expressions often involves various techniques for reducing algebraic expressions to their simplest form. For logarithmic expressions, simplification could involve utilizing identities and rules such as the power rule and inverse functions.
In our specific problem, after first using the power rule to transform \( \ln(e^{1/3}) \) to \( \frac{1}{3} \cdot \ln(e) \), we then applied our understanding of inverse functions. By recognizing that \( \ln(e) = 1 \), this allowed the expression \( \frac{1}{3} \cdot \ln(e) \) to simplify further to \( \frac{1}{3} \cdot 1 \), which results in \( \frac{1}{3} \).
When simplifying, it's crucial to identify and apply suitable properties and rules. Doing so not only shortens the process but also makes complex-looking expressions easier to manage and understand. This practice is fundamental in coursework involving calculus and algebra, especially when dealing with logarithmic and exponential functions.