Problem 72 - Algebra 2

Question

Find the inverse of each function. $$ y=\log _{a}(x-b) $$

Step-by-Step Solution

Answer

The inverse of the function y = \( \log_a {(x-b)} \) is \( y = a^x + b \)

Step 1: Swap x and y

To start with, swap the places of x and y in the equation to find the inverse. This will result in: x = \( \log_a {(y-b)} \).

Step 2: Transform into exponential form

Transform the equation into exponential form based on the property of logarithms. This results in \( a^x = y - b \).

Step 3: Solve for y

Solving for y will give you your inverse function. To get y by itself on one side of the equation, add b to both sides. This will result in: \( y = a^x + b \).

Key Concepts

Logarithmic Functions

Logarithmic functions are essential in mathematics as they serve as the inverses of exponential functions. When you have an expression like \(y = \log_a(x-b)\), it means you're asking the question: "To what power must the base \(a\) be raised to produce \(x-b\)?" Logarithms help us solve equations where the unknown appears as an exponent, making them extremely valuable in many areas of science and engineering.
The base \(a\) in a logarithmic function is crucial as it determines the growth behavior of the function. Common bases include 10 (common logarithm) and \(e\) (natural logarithm).
When dealing with logarithmic equations, one common strategy is to swap variables to find an inverse. This involves replacing the dependent variable with the independent one, an action that starts the process of finding an inverse function.

Exponential Form

Converting logarithmic equations into exponential form is a key skill necessary for solving these kinds of problems. When you see an equation like \(x = \log_a(y-b)\), you can convert it by recalling the relationship between logarithms and exponents.
In exponential form, the equation becomes \(a^x = y - b\). This translation works because logarithms and exponential functions are inverse operations; every logarithmic statement can be rewritten as an exponential statement and vice versa.

• To move from log form to exponential form, the base \(a\), along with the logarithm's output (which in our revised equation is \(x\)), serves as the inputs to the exponent operation to produce \(y - b\).

Understanding this transformation unlocks the techniques needed for further algebraic manipulation, helping to isolate variables and solve for them.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying mathematical expressions to achieve a desired form. In the context of finding the inverse of logarithmic functions, such as \(x = \log_a(y-b)\), algebraic manipulation facilitates isolating \(y\).
Once the equation is in exponential form, \(a^x = y - b\), the next step is to apply algebraic operations that will solve for \(y\):

• Begin by adding \(b\) to both sides to counter the subtraction, resulting in \(y = a^x + b\).
• Ensure the expression is simplified and makes logical sense within the given context. For instance, the equation should highlight \(y\) as a function of \(x\), reflecting the inverse relationship.

These algebraic skills are not just for solving equations but also help in graphing functions, hiding complexity, or revealing simpler relationships within a function. Mastering them aids in tackling a wide range of mathematical problems efficiently.