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Magazine Chapter 1: Problem 31
Determine at which points the graphs of the given pair of functions intersect. $$ f(x)=e^{3 x} \text { and } g(x)=3 e^{x} $$
Short Answer
Expert verified
The graphs intersect at \( \left( \frac{\ln(3)}{2}, 3^{3/2} \right) \).
Step by step solution
01
Set the Equations Equal
To find the points where the graphs intersect, set the two functions equal to each other: \[ e^{3x} = 3e^x \] This equation will help us find the values of \( x \) where the graphs intersect.
02
Simplify the Equation
Divide both sides of the equation \( e^{3x} = 3e^x \) by \( e^x \): \[ e^{3x}/e^x = 3e^x/e^x \] This simplifies to \[ e^{2x} = 3 \].
03
Solve for x
Take the natural logarithm of both sides of the equation \( e^{2x} = 3 \):\[ \ln(e^{2x}) = \ln(3) \]Since \( \ln(e^{2x}) = 2x \ln(e) = 2x \), we have:\[ 2x = \ln(3) \]Divide both sides by 2 to solve for \( x \):\[ x = \frac{\ln(3)}{2} \].
04
Find the Corresponding y-Value
Plug \( x = \frac{\ln(3)}{2} \) back into one of the original functions to find \( y \). We will use \( f(x) = e^{3x} \):\[ y = e^{3(\frac{\ln(3)}{2})} = e^{\frac{3\ln(3)}{2}} = (e^{\ln(3)})^{3/2} = 3^{3/2} \].Thus, the \( y \)-coordinate of the intersection point is \( 3^{3/2} \).
05
Conclusion
The two functions intersect at the point where \( x = \frac{\ln(3)}{2} \) and \( y = 3^{3/2} \), thus the point of intersection is \[ \left( \frac{\ln(3)}{2}, 3^{3/2} \right) \].
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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. In our exercise, both functions, \( f(x) = e^{3x} \) and \( g(x) = 3e^x \), are exponential. Here's what makes exponential functions unique:
- The constant \( e \) is approximately 2.718 and is also known as Euler's number. It's a critical constant in mathematics, especially in calculus.
- Exponential functions grow at rates proportional to their current value, which means they can model rapid increases or decreases.
- In our specific functions, \( e^{3x} \) means the base \( e \) is raised to a power that is three times \( x \). For \( 3e^x \), the expression implies a multiplication of the exponential part \( e^x \) with 3.
Solving Equations
Solving equations, especially those with exponential expressions, involves finding the value(s) of the unknown variable which satisfy the equation. Here, we aimed to solve \( e^{3x} = 3e^x \) to identify the intersection points of the graphs.
- The initial step in solving involved setting the two functions equal to find where they intersect. This produced \( e^{3x} = 3e^x \).
- To simplify, both sides of the equation were divided by \( e^x \), using the property of exponents that \( \frac{e^{3x}}{e^x} = e^{3x-x} = e^{2x} \).
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \). In solving the equation \( e^{2x} = 3 \), the natural logarithm helps to "unlock" the exponent:
- Taking \( \ln \) of both sides gives \( \ln(e^{2x}) = \ln(3) \).
- Using the logarithmic identity \( \ln(e^a) = a \), we simplify \( \ln(e^{2x}) \) to \( 2x \).
- This results in \( 2x = \ln(3) \), a much simpler linear equation.
Coordinate Geometry
Coordinate geometry involves the study of geometric figures through a coordinate system. In our exercise, the task was to determine where two functions intersect on the Cartesian plane.
- The solution process required finding the \( x \)-coordinate and the corresponding \( y \)-coordinate of the intersection.
- After finding \( x = \frac{\ln(3)}{2} \), we substituted back into one of the original functions, \( f(x) = e^{3x} \), to evaluate \( y \).
- The substitution calculated \( y = 3^{3/2} \), which provided us with the \( y \)-coordinate.
