Problem 93 - Algebra 2

Question

Which function is the inverse of \(f(x)=2^{x-1} ?\) $$ \begin{array}{ll}{\text { A. } f^{-1}(x)=\log _{2}(x-1)} & {\text { B. } f^{-1}(x)=\log _{2}(x+1)} \\ {\text { C. } f^{-1}(x)=\log _{2} x-1} & {\text { D. } f^{-1}(x)=\log _{2} x+1}\end{array} $$

Step-by-Step Solution

Answer

The inverse of \(f(x)=2^{x-1}\) is \(f^{-1}(x)=\log _{2} x+1\).

Step 1: Identify the original function

The original function is given as \(f(x) = 2^{x-1}\).Assume that the inverse function \(f^{-1}(x) = y\). Therefore the given function can be rewritten as \(y = 2^{x-1}\).

Step 2: Interchange the function’s inputs and outputs

To find the inverse function, switch the roles of \(y\) and \(x\). This step would change the equation from \(y = 2^{x-1}\) to \(x = 2^{y-1}\).

Step 3: Solve the equation for \(y\)

Isolate \(y\) on one side of the equation. Since the equation \(x = 2^{y-1}\) involves an exponent, use logarithms to 'undo' the exponent on the right side. Take the log base 2 of both sides to get \(\log_2{x} = \log_2{2^{y-1}}\). By the property of logarithms, if \(\log_b a = \log_b c\) then \(a = c\). So we now have \(\log_2{x} = y - 1\). Solving for \(y\), we get \(y = \log_2{x} + 1\).

Step 4: Identify the correct inverse function from the given options

Comparing the above derivation with the given options in the problem statement, the inverse function we found matches with Option D: \(f^{-1}(x)=\log _{2} x+1\).

Key Concepts

Logarithms

Logarithms are the mathematical tool we use to reverse exponential functions. They're essentially the opposite operation of exponentiation.
In simple terms, if you have a number and you're trying to figure out what power you raised another number to get it, you're dealing with a logarithm. For example, if you have 8 and you know it's a power of 2, using a logarithm helps you find that power, which is 3 (i.e., because \( 2^3 = 8 \)).

• Logarithms have a base, similar to exponents. Common bases are 10 (common logarithm) and \( e \) (natural logarithm), but in many problems, forming inverse operations with base 2, like here, is essential.
• An important property is: \( \log_b{a^c} = c \cdot \log_b{a} \).

Understanding logarithms is crucial for breaking down how exponential growth "works backward," as seen in this exercise to find inverse functions.

Exponential Functions

Exponential functions rapidly increase or decrease, depending on the scenario.
An example of an exponential function is \( f(x) = 2^{x-1} \). Here, "2" is the base, which indicates how quickly the function grows or decays.

• Exponential growth means as \( x \) increases, \( f(x) \) grows very fast. Conversely, if \( x \) decreases, the function value approaches zero.
• These functions are critical in modeling phenomena such as population growth or radioactive decay. They also play a key role in calculus and higher-level algebra.

To find an inverse, understanding that exponentiation needs to be "reversed" by a logarithm is key. In this exercise: finding \( f^{-1}(x) \) means "undoing" the exponent using logarithms.

Function Transformations

Function transformations help us understand how to modify and manipulate graphs.
By changing certain components of a function, we can shift, stretch, or compress graphs. For instance, the function \( f(x) = 2^{x-1} \) involves a transformation known as horizontal shifting.

• A horizontal shift occurs when you add or subtract a number in the exponent. In \( f(x) = 2^{x-1} \), subtracting 1 causes the graph to shift to the right by 1 unit.
• Transformations give flexibility in graph interpretation and function adjustments.

When deriving or graphing inverse functions, recognizing how transformations affect the parent function simplifies understanding its characteristics.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate variables, aiming to simplify or solve them.
For finding inverses, especially when exponents or logarithms are involved, step-by-step algebraic rearrangements are critical.

• In our exercise: beginning with \( x = 2^{y-1} \), algebraic steps require switching and solving for \( y \).
• Take the logarithm of both sides: \( \log_2{x} = y - 1 \).

Then, solving: add 1 to both sides to isolate \( y \), resulting in \( y = \log_2{x} + 1 \). This process was key to identifying the correct inverse function from the options provided in the exercise, demonstrating how essential algebraic manipulation is in mathematics.