Problem 72 - Precalculus Mathematics for Calculus

Question

Find the inverse function of \(f\). $$f(x)=3^{x+1}$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \log_3(x) - 1 \).

Step 1: Substitute and Interchange Variables

To find the inverse function, we first substitute the function output with a new variable and solve for the input variable. Let’s set the function to be equal to a new variable, say, \( y \). Thus, we have: \( y = 3^{x+1} \). Now, switch the variables \( x \) and \( y \) to prepare for solving: \( x = 3^{y+1} \).

Step 2: Isolate the Exponential Expression

To isolate \( y \), you need to express it in terms of \( x \). Start by applying logarithms to both sides. Since \( 3 \) is the base of the exponential, we will use base-3 logarithms. So, apply \( \log_3 \) on both sides: \( \log_3(x) = \log_3(3^{y+1}) \).

Step 3: Simplify Using Logarithmic Identities

By using the logarithmic identity \( \log_b(b^a) = a \), simplify the right side: \( \log_3(x) = y+1 \).

Step 4: Solve for \( y \)

Subtract 1 from both sides to isolate \( y \): \( y = \log_3(x) - 1 \).

Step 5: Conclude the Inverse Function

Thus, the inverse function \( f^{-1}(x) \) is given by swapping \( y \) back with \( x \): \( f^{-1}(x) = \log_3(x) - 1 \).

Key Concepts

Exponential Function

The exponential function is a mathematical function of the form \( f(x) = a^{bx+c} \). Here, \( a \) is the base of the exponential, \( b \) and \( c \) are constants, and \( x \) is the variable or exponent. In the context of our exercise, the function provided is \( f(x) = 3^{x+1} \). This means that our exponential function has a base of 3, and the exponent is \( x+1 \).

Exponential functions are powerful tools in mathematics as they describe growth and decay processes accurately, such as compound interest or population growth. These functions are characterized by a constant ratio of change, meaning they grow by a consistent factor over equal intervals.

• The graph of an exponential function is a curve that rises (for growth) or falls (for decay) rapidly, known as an exponential curve.
• The domain of an exponential function is all real numbers.
• The range is all positive numbers when the base is greater than 1.

Understanding exponential functions is essential when solving for inverse functions, as they often involve switching to logarithmic functions to isolate variables.

Logarithmic Identities

Logarithmic identities are mathematical tools that help us manipulate log expressions conveniently, facilitating the process of solving equations. The logarithm is essentially the inverse operation of exponentiation.

To better handle logarithmic expressions in solving for inverses, it's crucial to know these basic identities:

• \( \log_b(b^a) = a \) - This identity states that the logarithm of a base raised to a power simplifies to the exponent.
• \( \log_b(xy) = \log_b(x) + \log_b(y) \) - Useful for decomposing the logarithm of a product into a sum.
• \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \) - This identity is beneficial for handling logarithmic division.
• \( \log_b(x^n) = n\cdot\log_b(x) \) - It simplifies the management of powers within logs.

In our task of finding the inverse function, the identity \( \log_3(3^{y+1}) = y+1 \) played a central role after applying the base-3 logarithm on both sides. This simplification allowed us to isolate the variable \( y \) effectively.

Base-3 Logarithm

A base-3 logarithm, denoted as \( \log_3(x) \), is a logarithm with a base of 3. It asks the question: "To what power must 3 be raised to obtain \( x \)?". Using a base-3 logarithm is particularly useful in solving equations that involve powers of 3, primarily because it directly counters the exponential terms with the same base.

In solving the exercise, we applied \( \log_3 \) on both sides of the equation \( x = 3^{y+1} \). This conversion simplifies the exponential term and leads to a more straightforward algebraic expression \( \log_3(x) = y+1 \). By isolating \( y \) afterward, we determine the inverse function as \( y = \log_3(x) - 1 \).

This step highlights why choosing a logarithm with the matching base of the exponential function is effective:

• It allows for the direct simplification of power expressions.
• The overall calculation process becomes more intuitive.
• This method streamlines solving for inverses, as the conversion aligns perfectly with the structure of the exponential expression.

Mastering the relationship between exponential functions and their corresponding logarithms is crucial for tackling problems involving inverse functions.