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Exponential functions are a type of mathematical function where a constant base raised to a variable exponent. For example, in the function \(F(t) = e^{3t}\), the base is the constant \(e\) (Euler's number, approximately 2.718), and the exponent is the variable expression \(3t\).
Exponential functions have several key properties:

• They grow very quickly as the exponent increases.
• They have a horizontal asymptote at \(y = 0\), meaning they approach but never touch the x-axis as \(t\) goes to negative infinity.
• The derivative of an exponential function with base \(e\) is proportional to the function itself, i.e., \frac{\text{d}}{\text{d}x} e^{x} = e^{x}.

Understanding these properties is essential when analyzing and simplifying expressions within exponential functions.
In the given problem, function \(F(t) = e^{3t}\) represents an exponential function where the variable exponent is \(3t\). To find the inverse of an exponential function, you typically use a logarithmic function, which we will discuss next.

A logarithmic function is the inverse of an exponential function. It essentially means that if the exponential function represents growth, the logarithmic function represents the time or effort needed to achieve that level of growth. In simpler terms, while an exponential function of the form \(a^x = y\) tells us what \(y\) is for any given \(x\), the logarithmic function \(\text{log}_a(y) = x\) tells us what \(x\) needs to be to get \(y\).

For instance, the given function \(G(t) = \frac{1}{3} \text{ln}(t)\) involves the natural logarithm, which is the inverse function of \(e^x\). The natural logarithm \(\text{ln}(t)\) is specifically designed to reverse the process of exponentiation with the base \(e\). In the exercise solution, simplifying \(G(t) = \text{ln}(t^{1/3})\) correctly translates to \(G(t) = \frac{1}{3} \text{ln}(t)\).

Important properties of logarithmic functions include:

• \text{ln}(e^x) = x
• \text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b)
• \text{ln}(a^b) = b \times \text{ln}(a)

In the context of the problem, we utilize the property that \(\text{ln}(e^{3t}) = 3t\) to show that applying \(G\) to \(F\) returns the original input \(t\).

Function composition involves applying one function to the results of another function. Given two functions \(f\) and \(g\), the composition \(f(g(x))\) means you first apply \(g\) to \(x\), and then apply \(f\) to the outcome of \(g(x)\).

In this exercise, we need to prove that the functions \(F(t) = e^{3t}\) and \(G(t) = \frac{1}{3} \text{ln}(t)\) are inverses. This means that performing \(G\) on the output of \(F\), and vice versa, should return the original input \(t\). Let's break this down:

• Step 1: Compose \(G\) with \(F\): \(G(F(t)) = G(e^{3t})\). Substituting this into \(G\), we get \(\frac{1}{3} \text{ln}(e^{3t}) = t\).
• Step 2: Compose \(F\) with \(G\): \(F(G(t)) = F(\text{ln}(t^{1/3}))\). Substituting this into \(F\), we get \(e^{3 \times \frac{1}{3} \text{ln}(t)} = t\).

When both compositions return the input \(t\), it confirms that \(F\) and \(G\) are indeed inverse functions. Understanding function composition is crucial for verifying this property.