Question

Find the inverse function of \(y=\sigma b^{c t}\)

Short answer

The inverse function is \( t = f^{-1}(y) = \frac{\ln\left(\frac{y}{\sigma}\right)}{c \ln(b)} \).

Step-by-step solution

Understand the Inverse Function

To find the inverse function, we need to express the independent variable in terms of the dependent variable. In mathematical terms, we will rewrite the function so that we solve for the variable inside the function, which is here represented as \(t\) in terms of \(y\).

Start with the Original Function

Given the function \(y = \sigma b^{ct}\), it expresses \(y\) in terms of \(t\). To find the inverse, we'll rearrange and solve this equation for \(t\).

Isolate the Exponential

First, divide both sides by \(\sigma\) to get the term with the exponential function by itself: \(\frac{y}{\sigma} = b^{ct}\).

Apply the Logarithm

To solve for \(t\), apply the logarithm to both sides. You can use any logarithm, but we'll use the natural logarithm (ln): \[ \ln\left(\frac{y}{\sigma}\right) = \ln\left(b^{ct}\right) \]

Use the Power Rule of Logarithms

Utilize the power rule \(\ln(a^k) = k \cdot \ln(a)\) to bring down the exponent: \[ \ln\left(\frac{y}{\sigma}\right) = ct \cdot \ln(b) \]

Solve for t

Solve for \(t\) by dividing both sides of the equation by \(c \ln(b)\):\[ t = \frac{\ln\left(\frac{y}{\sigma}\right)}{c \ln(b)} \]

Write the Inverse Function

Now, express \(t\) as a function of \(y\). The inverse function will be:\[ t = f^{-1}(y) = \frac{\ln\left(\frac{y}{\sigma}\right)}{c \ln(b)} \]

Key concepts

Exponential Functions

In mathematics, exponential functions are critical for modeling situations where growth or decay happens at a constant rate. An exponential function retains its form such that one variable is raised to the power of another. In the original exercise, the function is expressed as \(y = \sigma b^{ct}\). Here, \(b\) represents the base of the exponential, which dictates how rapidly or slowly the function grows or decays, while \(ct\) demonstrates the exponent which incorporates time or another variable that influences growth.
The constant \(\sigma\) may represent an initial value or scaling factor, primarily affecting the range of the function but not the growth rate. Understanding the growth pattern of exponential functions is crucial, especially in contexts like population dynamics, radioactive decay, and financial interest calculations. They depict how quantities compound over successive periods, offering invaluable insights into various real-world applications.

• ### Key Characteristics:
  • Base \(b > 1\) causes a growth.
  • Base \(0 < b < 1\) causes a decay.
• ### Graph Behavior:
  • Always crossing the y-axis unless shifted.
  • Cannot take negative values in base \(b > 1\) scenarios.

Logarithmic Functions

Logarithmic functions can be envisioned as the inverse functions of exponential functions. If you have an exponential function where the independent variable is the exponent, a logarithmic function lets you solve for this variable. In simpler terms, logarithms tell you the power to which a base must be raised to get a certain number.
In the solution steps, to invert the exponential function \(y = \sigma b^{ct}\), we use logarithms to "unlock" the exponent \(ct\) by applying a logarithmic operation to both sides of the equation. Specifically, the natural logarithm (\(\ln\)) is applied, as it is the inverse of the natural base \(e\), a common mathematical constant.

• ### Solving Exponential Equations with Logarithms:
  • Separate the exponential expression by isolating it.
  • Apply logarithms—here the natural logarithm—on both sides.
  • Utilize the log power rule: \(\ln(a^k) = k \cdot \ln(a)\).
• ### Useful Properties:
  • Logarithms convert multiplicative processes into additive processes.
  • Help in determining time periods in compound interest and half-lives in radioactive decay.

Mathematical Problem Solving

Finding inverse functions is a vital skill in mathematical problem solving, especially within calculus and algebra. The goal here is to express one variable in terms of another, essentially reversing the function's effect. This requires rearranging and solving the original equation for the desired variable.
Critical to this process is a sound grasp of algebraic manipulations, such as isolating terms and applying logarithmic and exponential rules. In this exercise, when we isolated \(ct\) in the function \(y = \sigma b^{ct}\), we used algebraic steps of division and applied the natural logarithm to transform the equation into a solvable format.

• ### Steps in Problem Solving:
  • Start by understanding the problem and identifying the function to be inverted.
  • Isolate the exponential term by removing any coefficients or divisors.
  • Apply the appropriate inverse operations—here, using logarithms.
• ### Techniques Employed:
  • Isolating the variable: remove constants and coefficients.
  • Applying inverse functions: logarithms for exponential functions.

Possessing these strategies enriches your mathematical toolkit, making it easier to tackle other complex problems across various math topics.