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Chapter 5: Problem 41

The supply function and the demand function for the sale of a certain type of DVD player are given by $$S(p)=150 e^{0.004 p} \quad \text { and } \quad D(p)=480 e^{-0.003 p}$$ respectively, where \(S(p)\) is the number of DVD players that the company is willing to sell at price \(p\) and \(D(p)\) is the quantity that the public is willing to buy at price \(p\). Find \(p\) such that \(D(p)=S(p)\) This is called the equilibrium price.

Short Answer

Expert verified
The equilibrium price is approximately 166.16.

Step by step solution

01

- Set Demand Equal to Supply

To find the equilibrium price, set the demand function equal to the supply function: \[ D(p) = S(p) \]. Using the given functions, this translates to: \[ 150e^{0.004p} = 480e^{-0.003p} \].
02

- Simplify the Equation

Divide both sides of the equation by 150 to simplify: \[ e^{0.004p} = \frac{480}{150}e^{-0.003p} \]. Simplify the fraction: \[ e^{0.004p} = 3.2 e^{-0.003p} \].
03

- Combine the Exponents

Rewrite the equation to combine the exponents into one term: \[ e^{0.004p + 0.003p} = 3.2 \]. This simplifies to: \[ e^{0.007p} = 3.2 \].
04

- Solve for p using Natural Logarithm

Take the natural logarithm on both sides to solve for p: \[ \ln(e^{0.007p}) = \ln(3.2) \]. Using the property of logarithms \(\ln(e^x) = x\), we get: \[ 0.007p = \ln(3.2) \].
05

- Isolate p

Divide both sides by 0.007 to isolate p and find its value: \[ p = \frac{\ln(3.2)}{0.007} \]. Using a calculator, \ln(3.2) \approx 1.16315, so: \[ p \approx \frac{1.16315}{0.007} \approx 166.16 \].

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Key Concepts

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

supply and demand
Supply and demand is a fundamental concept in economics. It describes the relationship between the amount of a good or service that producers are willing to sell and the amount that consumers are willing to buy.

At any given price, producers have a certain quantity they will supply (based on their costs, desired profit, etc.) and consumers have a quantity they will demand (based on their preferences, income, etc.).
  • Supply function: Represents how much of a good will be supplied at different prices.
  • Demand function: Represents how much of a good will be demanded at different prices.
The equilibrium price is the price at which the quantity supplied equals the quantity demanded.

This is the point where supply meets demand, and both consumers and producers are satisfied with the price and quantity.
natural logarithm
The natural logarithm is a mathematical function often denoted by \(\text{ln}(x)\). It has several important properties and is used extensively in calculus and in various applications in science and engineering.

The natural logarithm is the inverse of the exponential function with base \(e\), where \(e\) is approximately equal to 2.71828. Here are some key points:
  • If \(e^y = x\), then \(y = \text{ln}(x)\).
  • The natural logarithm of \(1\) is \(0\): \(\text{ln}(1) = 0\).
  • The natural logarithm of \(e\) is \(1\): \(\text{ln}(e) = 1\).
In this exercise, we use the natural logarithm to solve the equation involving exponential terms. Specifically, we took the natural logarithm of both sides of the equation \(\text{e}^{0.007p} = 3.2\) to isolate \(p\).
exponential equations
Exponential equations are equations in which variables appear as exponents. Solving these equations requires understanding properties of exponents and logarithms.

In the given exercise, we started with two exponents: \(150e^{0.004p} = 480e^{-0.003p}\). Here are the general steps to solve:
  • First, simplify the equation so that each side has a single exponential term.
  • Combine or balance the exponents as needed.
  • Use logarithms to isolate the variable exponent.
By applying these steps, we were able to rewrite our equation to \(\text{e}^{0.007p} = 3.2\), simplifying the problem significantly. Taking the natural logarithm of both sides allowed us to solve for \(p\).
logarithmic properties
Logarithms have several important properties that make them useful for solving equations involving exponents. Some key properties include:
  • \(\text{log}_b(xy) = \text{log}_b(x) + \text{log}_b(y)\): The logarithm of a product is the sum of the logarithms.
  • \(\text{log}_b\big(\frac{x}{y}\big) = \text{log}_b(x) - \text{log}_b(y)\): The logarithm of a quotient is the difference of the logarithms.
  • \(\text{log}_b(x^y) = y \text{log}_b(x)\): The logarithm of a power is the exponent times the logarithm of the base.
  • \(\text{ln}(e^x) = x\): The natural logarithm of an exponential function is the exponent itself.
In this exercise, we used the property \(\text{ln}(e^x) = x\) to simplify and solve the equation. This property allowed us to convert an exponential equation into a linear form which was easier to handle and isolate \(p\). Without these properties, solving such equations would be much more difficult.

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