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Chapter 5: Problem 69

You are given the coordinates of two points on the graph of the curve \(y=a e^{b x} .\) In each case, determine the values of a and b. $$(-2,324) \text { and }(1 / 2,4 / 3)$$

Short Answer

Expert verified
\(a = 36.75\), \(b \approx -1.609\).

Step by step solution

01

Equation Setup for First Point

The given exponential curve is in the form \( y = a e^{bx} \). We have the point \((-2, 324)\), so substituting this into the equation, we get: \[324 = a e^{-2b} \].
02

Equation Setup for Second Point

Substitute the second point \((1/2, 4/3)\) into the equation \( y = a e^{bx} \):\[\frac{4}{3} = a e^{\frac{b}{2}} \].
03

Solving the System of Equations

We have the two equations now:1. \(324 = a e^{-2b}\) 2. \(\frac{4}{3} = a e^{\frac{b}{2}}\) Divide the second equation by the first to eliminate \(a\):\[\frac{4/3}{324} = \frac{a e^{b/2}}{a e^{-2b}} \Rightarrow \frac{4}{3} \times \frac{1}{324} = e^{5b/2}\] \[e^{5b/2} = \frac{1}{243}\].
04

Solving for b

Using \(e^{5b/2} = \frac{1}{243}\), take the natural logarithm of both sides:\[\frac{5b}{2} = \ln\left(\frac{1}{243}\right)\]\[b = \frac{2}{5} \cdot (-\ln(243))\] \[b = -\frac{2}{5} \cdot \ln(243)\].
05

Solving for a

Using the value of \(b\) and substituting back into one of the original equations, \(324 = a e^{-2b}\): \[ a = \frac{324}{e^{-2b}} = 324\cdot e^{2b}\].Substitute \(b = -\frac{2}{5} \cdot \ln(243)\) to solve for \(a\):\[ a = 324 \times \left(e^{-\frac{4}{5} \ln(243)}\right) \].
06

Simplification of a and b

Calculate \(b\) numerically using \(b = -\frac{2}{5} \cdot \ln(243)\) and substitute this value into the equation for \(a\). This will give the numerical values of \(a\) and \(b\).

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Key Concepts

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

Solving Systems of Equations
When faced with multiple equations to solve simultaneously, you're dealing with a system of equations. This occurs often in mathematics, especially when trying to find unknown values that satisfy multiple conditions or relationships. The goal is to find the values of the variables that make all equations in the system true at the same time.

In this exercise, you have two equations derived from two different points on an exponential curve:
  • Equation 1: \( 324 = a e^{-2b} \)
  • Equation 2: \( \frac{4}{3} = a e^{\frac{b}{2}} \)
To solve for \(a\) and \(b\), you can use the method of substitution or elimination. Here, dividing one equation by the other is a clever way to eliminate \(a\) and solve for \(b\). This approach simplifies the problem, making it easier to find the unknown parameters.
Coordinate Geometry
Coordinate geometry uses coordinate points to analyze geometric shapes and different curves. In analyzing the exponential curve \(y = a e^{bx}\), two given points, \((-2,324)\) and \(\left(\frac{1}{2}, \frac{4}{3}\right)\), help us understand the position and shape of the graph in the coordinate plane.

By substituting these points into the curve's equation, you develop a system of equations that connects the geometric model (graph) to algebraic analysis. Geometry helps visualize how these equations manifest in space, connecting the dots, quite literally, on where these solutions likely lie. This representation transforms abstract numbers into meaningful, visual information that aids in comprehending how the variables relate to each other on the curve.
Natural Logarithms
Natural logarithms are a fundamental mathematical concept helpful for solving equations involving exponentials, especially those in the form \(y = a e^{bx}\). The natural logarithm, denoted by \(\ln\), is the inverse of the exponential function, making it invaluable for solving equations where the unknown is an exponent.

In the context of this problem, after obtaining \(e^{\frac{5b}{2}} = \frac{1}{243}\), taking the natural logarithm of both sides is essential. This converts the equation to linear form, making it easier to solve for \(b\). You use the property that \(\ln(e^x) = x\), which simplifies the process dramatically. Understanding and applying natural logarithms effectively is crucial in solving many problems involving exponential functions.

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Most popular questions from this chapter

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$4\left(10-e^{x}\right) \leq-3$$

The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.

Estimate a value for \(x\) such that \(\log _{2} x=100 .\) Use the approximation \(10^{3}=2^{10}\) to express your answer as a power of 10 Answer: \(10^{30}\)

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.
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