Chapter 6: Problem 76
Find the points on the graph of the given function that have the indicated \(y\) -coordinate. $$ f(x)=25^{x}-5^{x+1} ;-6 $$
Short Answer
Expert verified
The points are \( (\log_5(2), -6) \) and \( (\log_5(3), -6) \).
Step by step solution
01
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: \( 25^x - 5^{x+1} = -6 \).
02
Simplify the Expression
First, recognize that \( 25^x = (5^2)^x = (5^x)^2 \). Rewriting the expression, our equation becomes: \((5^x)^2 - 5^{x+1} = -6\).
03
Use Substitution
To simplify further, let \( u = 5^x \). Then the equation becomes: \( u^2 - 5u = -6 \).
04
Solve the Quadratic Equation
This equation \( u^2 - 5u + 6 = 0 \) is a quadratic in standard form. Factor the quadratic as \((u-2)(u-3) = 0\).
05
Find the Values of u
From the factored form, set \( u-2 = 0 \) and \( u-3 = 0 \). Thus, \( u = 2 \) or \( u = 3 \).
06
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 \( 5^x = 3 \), \( x = \log_5(3) \).
07
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 \( (\log_5(3), -6) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( u = 3 \).
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 \( x \):
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 \( x = \log_5(3) \).