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Exponents are a fundamental concept in mathematics. They represent how many times a number, known as the base, is multiplied by itself. For instance, in the expression \( 3^2 \), the base is 3, and the exponent is 2. This expression means 3 multiplied by itself, which equals 9.

When dealing with exponents, there are a few important rules to remember:

• Any number raised to the power of 1 is itself: \( a^1 = a \).
• Any number raised to the power of 0 equals 1: \( a^0 = 1 \), assuming \( a eq 0 \).
• For any positive integer n, \( a^{-n} = {1}/{a^n} \). For example, \( 3^{-2} = {1}/{3^2} = {1}/{9} \).

Understanding exponents is crucial for manipulating logarithmic functions, as they are the inverse of each other.

Inverse functions are functions that 'undo' each other. If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), then applying \( f \) and then \( f^{-1} \) will bring you back to your original value. Mathematically, this is expressed as: \( f(f^{-1}(x)) = x \).

In the context of logarithms and exponents, the logarithm function is the inverse of the exponential function. For example, if \( f(x) = 3^x \), then \( f^{-1}(x) = \log_3(x) \), meaning that exponentiating a number undoes the logarithm and vice versa.

With our given function \( f(x) \), where \( f(x) \) is the exponent that gives us a certain number with base 3, we are essentially working with the logarithmic function base 3: \( \log_3(x) \). This inversion principle is the core idea behind finding values in logarithmic functions such as base-3 logarithms.

Base-3 logarithms are a specific type of logarithm where the base is 3. They tell us the exponent needed to raise 3 to get a particular number. The notation is: \( \log_3(x) \).

Here are a few examples to clarify:

• \( \log_3(27) = 3 \) since \( 3^3 = 27 \).
• \( \log_3(81) = 4 \) since \( 3^4 = 81 \).
• \( \log_3(3) = 1 \) since \( 3^1 = 3 \).

These examples show the application of the logarithm function. By understanding these, you can solve logarithmic equations more effectively.

It is also beneficial to recognize when dealing with fractions. For example, \( \log_3(\frac{1}{9}) \): Since \( 3^{-2} = \frac{1}{9} \), it follows that \( \log_3(\frac{1}{9}) = -2 \).

Evaluating functions involves finding the value of a function at a particular input. For the original exercise, we evaluate the function \( f(x) \) for different values of x by determining the exponent needed to produce that value when using 3 as the base:

• For \( f(27) \), we find the exponent a such that \( 3^a = 27 \). Since \( 3^3 = 27 \), \( f(27) = 3 \).
• For \( f(81) \), we find the exponent a such that \( 3^a = 81 \). Since \( 3^4 = 81 \), \( f(81) = 4 \).
• For \( f(3) \), we find the exponent a such that \( 3^a = 3 \). Since \( 3^1 = 3 \), \( f(3) = 1 \).
• For \( f(\frac{1}{9}) \), we find the exponent a such that \( 3^a = \frac{1}{9} \). Since \( 3^{-2} = \frac{1}{9} \), \( f(\frac{1}{9}) = -2 \).

Evaluating functions like this helps in understanding how the values of functions are determined and applied in real-world problems.