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Chapter 5: Problem 75

Approximate the point \((s)\) of intersection of the pair of equations. $$y=2.3 \ln (x+10.7), y=10 e^{-0.007 x^{2}}$$

Short Answer

Expert verified
The x-coordinate of the intersection is approximately x = -6.

Step by step solution

01

- Understand the given equations

The given equations are: 1. \(y=2.3 \ln (x+10.7)\) 2. \(y=10 e^{-0.007 x^{2}}\)
02

- Set the equations equal to each other

Since both equations equal y, set them equal to each other to find the x-coordinate of intersection: \[2.3 \ln (x + 10.7) = 10 e^{-0.007 x^{2}}\]
03

- Approximate the solution graphically

Plot both equations on the same graph and find the intersection point. A graphing calculator or software can be used for this.
04

- Verify intersection point

After plotting, check the approximate x-value found by substituting back into both equations to ensure both give the same y-value.
05

- Solution approximation

From the graph, estimate the point of intersection. For these specific equations, the x-coordinate of the intersection point is around x = -6.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

logarithmic functions
A logarithmic function is the inverse of an exponential function. In this exercise, we have the logarithmic function defined as: \(y = 2.3 \, \text{ln}(x + 10.7)\).
Here, the logarithm \ provides the power to which a fixed number (base) must be raised to make a number. The base of the natural logarithm (ln) is e (approximately 2.71828).
Key properties of logarithmic functions include:
  • They pass through the point (1, 0) because \ \(ln(e^0) = 0\).
  • The domain is limited to positive real numbers - for \( \text{ln}(x + 10.7)\), we need \( x + 10.7 > 0\).
  • Logarithmic functions grow slower than any polynomial function.

Here, as x increases, the value of \( \text{ln}(x + 10.7) \) also increases, but at a decreasing rate.
exponential functions
Exponential functions involve constants raised to variable exponents. In this case, the function given is: \( y = 10e^{-0.007 x^{2}} \).
An exponential function can model scenarios where growth or decay accelerates (or decelerates) over time.
Important aspects of exponential functions include:
  • Functions of the form \( y = ae^{bx} \) where e is Euler's number (approximately 2.71828).
  • An exponential function grows (for positive b) or decays (for negative b) rapidly.
  • They are always positive for all real values of x because \( e^x \) is never zero.

This specific example represents a decay because of the negative exponent \( -0.007x^2 \). Hence, as x increases, the value of \( e^{-0.007x^2} \) decreases, making the curve rapidly approach zero.
graphing equations
Graphing equations helps visualize their behavior and find intersecting points. To solve this exercise, we plot: \( y = 2.3 \, \text{ln}(x + 10.7) \) and \( y = 10e^{-0.007 x^{2}} \) on a coordinate plane.
Here's how to proceed:
  • Use a graphing calculator or software like Desmos or GeoGebra.
  • Plot both functions over an appropriate range of x values.
  • Examine where the graphs intersect鈥攖his point is where both equations have the same y for the same x.

This visual approach aligns with the step of the original solution where graphing the equations helped approximate the intersection point at around \( x = -6 \).
approximating solutions
Approximating solutions involves estimating where functions intersect, especially when finding exact algebraic solutions is complex. Given the equations: \( y=2.3 \, \text{ln}(x+10.7) \) and \( y=10e^{-0.007 x^{2}} \), we can approximate the solution by graphing.
Steps to approximate solutions:
  • Plot both functions to see where their curves overlap.
  • Find the x-value where these functions appear to intersect.
  • Verify by substituting this x-value back into both equations to check if they yield the same y-value.

From our visual approximation, the point of intersection was around \( x = -6 \). Once you identify this value, you can refine your estimate by checking values closer to it to ensure greater accuracy.

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Most popular questions from this chapter

Advertising. A company begins an Internet advertising campaign to market a new telephone. The percentage of the target market that buys a product is generally a function of the length of the advertising campaign. The estimated percentage is given by $$ f(t)=100\left(1-e^{-0.04 t}\right) $$ where \(t\) is the number of days of the campaign. a) Graph the function. b) Find \(f(25),\) the percentage of the target market that has bought the phone after a 25 -day advertising campaign. c) After how long will \(90 \%\) of the target market have bought the phone?

Sketch the graph of the function and check the graph with a graphing calculator. Describe how each graph can be obtained from the graph of a basic exponential function. $$f(x)=1-e^{-0.01 x}$$

Solve using any method. Given that \(a=\log _{8} 225\) and \(b=\log _{2} 15,\) express as a function of \(b\).

Solve using any method. $$\left|2^{x^{2}}-8\right|=3$$

Suppose that \(\log _{a} x=2 .\) Find each of the following. Simplify: $$\log _{10} 11 \cdot \log _{11} 12 \cdot \log _{12} 13 \cdots \log _{998} 999 \cdot \log _{999} 1000$$
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