Problem 68 - Algebra 2

Question

Write an exponential equation \(y=a b^{x}\) whose graph passes through the given points. \(\left(-1, \frac{2}{3}\right)\) and \((2,18)\)

Step-by-Step Solution

Answer

The exponential function that passes through the points \((-1, 2/3)\) and \((2,18)\) is \(y = \frac{2}{9} \cdot 3^{x}\).

Step 1: Plug the points into the equation

Given an exponential function of the form \(y = a \cdot b^{x}\), stating that the points lie on the function's line means that if we input the \(x\)-coordinate from each point into the equation, the output should be the \(y\)-coordinate. Now let's plug in our two points into the equation: \((-1, 2/3)\) gives \(2/3 = a \cdot b^{-1}\) and \((2,18)\) gives \(18 = a \cdot b^{2}\). This generates a system of two equations to solve for our two unknowns, \(a\) and \(b\).

Step 2: Use the two equations to eliminate one variable

Our best option here is to solve one of the equations for \(a\) and then substitute that into the other equation to find \(b\). From the first equation, we have \(a = \frac{2}{3b}\). We substitute this into the second equation, which gives: \(18 = \frac{2}{3b} \cdot b^{2}\), which simplifies to \(18 = \frac{2b^{2}}{3b}\) and following through, we get \(b^3 = 27\).

Step 3: Solve for \(b\)

Taking the cube root on both sides of the equation \(b^3 = 27\) yields \(b = 3\).

Step 4: Substitute \(b\) into one equation to find \(a\)

Now that we have \(b = 3\), we can substitute it back into one of our original equations to find \(a\). Using the first equation, we get \(a = \frac{2}{3 \cdot 3}\), which simplifies to \(a = \frac{2}{9}\).

Key Concepts

Exponential Equation

An exponential equation is a mathematical expression where a constant base is raised to a variable exponent. They take the form of \( y = a \cdot b^{x} \), where \( a \) is a constant, \( b \) is the base that is greater than zero, and \( x \) is the exponent. Exponential functions are prevalent in growth and decay processes, like population growth and radioactive decay. The value of \( b \) determines the nature of the growth:

• If \( b > 1 \), the function represents exponential growth.
• If \( 0 < b < 1 \), it represents exponential decay.

When solving exponential equations, our goal is to find the values of the variables that satisfy the equation for given points on the curve. For example, given the points \((-1, \frac{2}{3})\) and \((2, 18)\), plugging these into the exponential equation form allows you to create equations that describe the relationship between variables in the equation.

System of Equations

A system of equations contains two or more equations with common variables. Solving these systems involves finding values that satisfy all equations simultaneously. In this exercise, we derive a system from the exponential form \( y = a \cdot b^{x} \) using the points \((-1, \frac{2}{3})\) and \((2,18)\):

• For point \((-1, \frac{2}{3})\), the equation is \( \frac{2}{3} = a \cdot b^{-1} \).
• For point \((2,18)\), the equation is \( 18 = a \cdot b^{2} \).

By solving this system, our goal is to determine the values of \( a \) and \( b \) that satisfy both equations. Solving such systems often involves a technique called substitution or elimination to reduce the number of variables and simplify the problem. In this example, once we express \( a \) in terms of \( b \) from one equation, we substitute back to solve for \( b \).

Cube Root

The cube root function is the inverse operation to cubing a number. The cube root of a number \( x \) is a number \( y \) such that when you multiply \( y \) by itself three times, you get \( x \). Mathematically, this is expressed as \( y^3 = x \) or \( y = \sqrt[3]{x} \).
In the context of the solution provided, we encountered the equation \( b^3 = 27 \). To find \( b \), we apply the cube root to both sides to obtain \( b = \sqrt[3]{27} \). Since 27 is \( 3^3 \), the cube root of 27 is indeed 3, giving us the solution \( b = 3 \). Understanding cube roots is essential when working with equations where the variable is cubed, especially in scenarios involving geometric or algebraic symmetry.

Solving for Variables

When solving for variables in mathematical equations, we aim to isolate the unknowns on one side of the equation to determine their values. In this lesson, we start with a system of equations derived from an exponential function. Through algebraic manipulation, we target variables \( a \) and \( b \).
The key steps include:

• Express one variable in terms of another, which can be done by rearranging one equation, such as \( a = \frac{2}{3b} \).
• Substitute it back into the second equation to eliminate one variable and solve for the remaining one, leading us to \( b^3 = 27 \).

After finding \( b \), substitute it back into the rearranged equation to solve for \( a \). This process demonstrates both substitution and logical reasoning in performing step-by-step manipulations, critical skills for mastering algebraic solutions.