Problem 76 - Precalculus with Calculus Previews

Question

Find the points on the graph of the given function that have the indicated \(y\) -coordinate. $$ f(x)=25^{x}-5^{x+1} ;-6 $$

Step-by-Step Solution

Answer

The points are \( (\log_5(2), -6) \) and \( (\log_5(3), -6) \).

Step 1: Set Up the Equation

We are given the function \( f(x) = 25^x - 5^{x+1} \) and need to find the value of \( x \) when the function equals \(-6\). Thus, we set up the equation: \( 25^x - 5^{x+1} = -6 \).

Step 2: Simplify the Expression

First, recognize that \( 25^x = (5^2)^x = (5^x)^2 \). Rewriting the expression, our equation becomes: \((5^x)^2 - 5^{x+1} = -6\).

Step 3: Use Substitution

To simplify further, let \( u = 5^x \). Then the equation becomes: \( u^2 - 5u = -6 \).

Step 4: Solve the Quadratic Equation

This equation \( u^2 - 5u + 6 = 0 \) is a quadratic in standard form. Factor the quadratic as \((u-2)(u-3) = 0\).

Step 5: Find the Values of u

From the factored form, set \( u-2 = 0 \) and \( u-3 = 0 \). Thus, \( u = 2 \) or \( u = 3 \).

Step 6: Back-substitute for x

Recall \( u = 5^x \). Therefore, solve \( 5^x = 2 \) and \( 5^x = 3 \) by taking the logarithm of both sides. For \( 5^x = 2 \), \( x = \log_5(2) \); for \( 5^x = 3 \), \( x = \log_5(3) \).

Step 7: Find Each Point on the Graph

Since there are two solutions for \( x \), the points corresponding to the given \( y \)-coordinate \(-6\) are \( (\log_5(2), -6) \) and \( (\log_5(3), -6) \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, such as in the function \( f(x) = a^x \), where \( a \) is a positive real number different from 1. In our exercise, the base is 25 and 5, leading to terms like \( 25^x \) and \( 5^{x+1} \). These functions are known for rapid growth or decay, depending on the base and the exponent's sign.

• When the base \( a \) is greater than 1, like in \( 25^x \), the function grows exponentially.
• If the base is a fraction between 0 and 1, the function decreases exponentially.

Recognizing that \( 25^x \) can be rewritten as \((5^2)^x\), which simplifies to \((5^x)^2\), is crucial in solving exponential equations. This simplification is an example of using mathematical properties to transform complex expressions into more manageable forms. Understanding these principles helps in breaking down and solving equations involving exponential terms by substituting simpler expressions.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). These equations often appear when you factor expressions or solve them through various methods like completing the square or using the quadratic formula. In our exercise, we encounter a transformed quadratic equation when substituting \( u = 5^x \). Our equation \( u^2 - 5u + 6 = 0 \) is derived from simplifying the original function.

• This specific quadratic was solved by factoring, a method where you express it as a product of binomials: \((u-2)(u-3) = 0\).
• By setting each factor equal to zero, \( u-2 = 0 \) and \( u-3 = 0 \), we find the solutions \( u = 2 \) and \( u = 3 \).

Quadratic equations often have two real solutions like in this case, showcasing the parabolic nature of their graphs. Analyzing these solutions helps us backtrack to our original variable \( x \), providing insights into the problem's requirements and helping identify specific points on graphs.

Logarithms

Logarithms are the inverse operation to exponentiation and are essential in solving equations where the variable is an exponent. They answer the question: "To what exponent must a base be raised, to yield a certain number?" In mathematical terms, \( a^x = b \) is equivalent to \( x = \log_a(b) \).
In our exercise, once we have \( u = 5^x \) set to known values like 2 and 3, we use logarithms to solve for \( x \):

• The equation \( 5^x = 2 \) becomes \( x = \log_5(2) \).
• For \( 5^x = 3 \), it transforms to \( x = \log_5(3) \).

These transformations are crucial for interpreting exponential relationships and finding exact values of variables that correspond to particular points on a graph. Understanding logarithms extends your ability to analyze and solve equations that involve exponential terms, making them a vital tool in both mathematics and real-world applications.