Problem 31 - College Algebra with Modeling and Visualization

Question

Find \(C\) and a so that \(f(x)=C a^{x}\) satisfies the given conditions. $$ f(1)=9 \text { and } f(2)=27 $$

Step-by-Step Solution

Answer

\(C = 3\) and \(a = 3\).

Step 1: Setting up Equations

Given that \(f(x) = C a^x\), we know from the problem that \(f(1) = 9\) and \(f(2) = 27\). This gives us two equations: \(C a^1 = 9\) and \(C a^2 = 27\).

Step 2: Express 'C' in Terms of 'a'

From the first equation, \(C a = 9\). We can express \(C\) as \(C = \frac{9}{a}\).

Step 3: Substituting 'C' into the Second Equation

Substitute \(C = \frac{9}{a}\) into the second equation, \(C a^2 = 27\). This gives \(\frac{9}{a} \cdot a^2 = 27\). Simplifying, we have \(9a = 27\).

Step 4: Solving for 'a'

From \(9a = 27\), divide both sides by 9 to get \(a = 3\).

Step 5: Solving for 'C'

Substitute \(a = 3\) back into \(C = \frac{9}{a}\). So, \(C = \frac{9}{3} = 3\).

Step 6: Verification of Solution

Verify the solution by checking both conditions: \(f(1) = 3 \cdot 3^1 = 9\) and \(f(2) = 3 \cdot 3^2 = 27\), which satisfy the given conditions.

Key Concepts

Understanding Algebra in Exponential Functions

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. In this problem, we are dealing with an exponential function, which is represented as \(f(x) = C a^x\). In an exponential function:

• \(C\) is called the coefficient. It affects the vertical stretch of the graph.
• \(a\) is the base of the exponent, which determines how fast the function grows or decays as \(x\) changes.

These two elements are crucial when you work with exponential functions, especially when you need to solve for specific values like in the given problem. Algebra helps us structure this function by setting up relationships between \(C, a,\) and other variables. In the context of this exercise, algebraic manipulation allows us to form equations like \(C a = 9\) and \(C a^2 = 27\), which we can use to solve for unknowns.

The Art of Solving Equations

Solving equations is about finding the values of missing variables that make the equation true. In this problem, we need to solve for both \(C\) and \(a\) using the equations derived from the conditions \(f(1) = 9\) and \(f(2) = 27\). The steps taken illustrate key solving techniques:

• **Formulate Equations:** Start with known conditions to form equations. Here, we have \(C a^1 = 9\) and \(C a^2 = 27\).
• **Substitution Method:** Solve one equation for a variable and substitute it into another. We first express \(C\) as \(C = \frac{9}{a}\) and substitute it into the next equation.
• **Simplification:** Simplify the equation to solve for one variable. After substitution, simplifying leads us to \(9a = 27\), and solving this gives \(a = 3\).
• **Verification:** Once values are found, verify by substituting them back into the original conditions to ensure they fit both equations.

These steps demonstrate logical thinking and systematic approaches to finding unknowns, which are essential skills in algebra and math.

Function Notation and Interpretation

Function notation is a way to express mathematical functions clearly and concisely. In this problem, \(f(x) = C a^x\) is the notation that describes the relationship between input \(x\) and output \(f(x)\). It's important to understand what each part of this notation represents:

• \(f(x)\) denotes the value of the function for a particular \(x\).
• \(C a^x\) represents how the function calculates this value by applying the exponential growth or decay based on \(a\) raised to the power \(x\) and scaled by \(C\).

This notation helps quickly communicate complex relationships without lengthy descriptions. Proper use of function notation also aids in easily substituting values. For example, you can directly compute \(f(1)\) and \(f(2)\) by plugging 1 and 2 into the function to find \(f(x)\) values, if you know \(C\) and \(a\). Understanding these notations makes it easier to handle complex problems and efficiently solve equations in algebra.