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Chapter 5: Problem 17

A pollutant was dumped into a lake, and each year its amount in the lake is reduced by \(25 \%\). a. Construct a general formula to describe the amount of pollutant after \(n\) years if the original amount is \(A_{0}\). b. How long will it take before the pollution is reduced to below \(1 \%\) of its original level? Justify your answer.

Short Answer

Expert verified
The general formula is \(A_n = A_0 \times (0.75)^n\). It will take approximately 17 years for the pollution to be reduced to below 1% of its original level.

Step by step solution

01

- Understanding the Problem

The problem gives the rate at which a pollutant decreases in a lake: 25% per year. The goal is to find a general formula to describe the amount of pollutant after n years, and then determine when the amount is less than 1% of its original level.
02

- Setting Up the General Formula

The original amount of pollutant is denoted as \(A_0\). Since 25% of the pollutant is reduced each year, 75% of the pollutant remains. Thus, after one year, the amount of pollutant is \(0.75A_0\). After two years, it is \((0.75)^2A_0\). This leads us to a general formula: \[ A_n = A_0 \times (0.75)^n \] where \(A_n\) is the amount of pollutant after n years.
03

- Finding When Pollution is Below 1%

To find when the pollution is reduced to below 1% of its original level, we set up the inequality: \[ A_0 \times (0.75)^n < 0.01A_0 \] Cancelling out \(A_0\) from both sides, we get: \[ (0.75)^n < 0.01 \]
04

- Solving the Inequality

Taking the natural logarithm of both sides: \[ \text{ln}((0.75)^n) < \text{ln}(0.01) \] Using the logarithm power rule, this simplifies to: \[ n \times \text{ln}(0.75) < \text{ln}(0.01) \] Since \( \text{ln}(0.75) \) is negative, we divide both sides by \( \text{ln}(0.75) \) and remember to reverse the inequality: \[ n > \frac{\text{ln}(0.01)}{\text{ln}(0.75)} \] Calculating this gives: \[ n > \frac{-4.605}{-0.288} \ n > 16 \ n \text{ is approximately } 17 \text{ years} \]

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Key Concepts

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

Pollutant Reduction
Pollutant reduction refers to the process by which the amount of a harmful substance decreases over time. In this scenario, 25% of the pollutant in a lake is reduced each year. This means that every year, a lesser amount of pollutant remains as compared to the previous year.
  • After one year, 75% of the pollutant remains.
  • After two years, it's 75% of what remained, and so on.
The continuous reduction creates a pattern that can be described by an exponential function.
Exponential Function
An exponential function models situations where a quantity changes by a constant percentage over equal increments of time. In our example, the pollutant reduces by 25% annually.
The formula: \[ A_n = A_0 \times (0.75)^n \] describes how the amount of pollutant ( \[ A_n \] ), changes after \[ n \] years, given the original amount ( \[ A_0 \] ). Here, 0.75 represents the remaining 75% of the pollutant each year.
This approach helps us to predict the future concentration of the pollutant more accurately.
Natural Logarithm
To solve equations involving exponential functions, we often use natural logarithms (ln). The natural logarithm helps us rearrange and solve equations where the unknown variable is an exponent.
In the exercise, we need to figure out when the pollutant level is less than 1% of its original amount. We set up the inequality: \[ A_0 \times (0.75)^n < 0.01A_0 \] By taking the natural logarithm of both sides, the exponential relationship is simplified: \[ \ln((0.75)^n) < \ln(0.01) \] Using the properties of logarithms, we then isolate \[ n \] to solve the inequality.
Inequality Solving
Inequality solving involves finding the values that satisfy an inequality. In our problem, the inequality is: \[ (0.75)^n < 0.01 \] After applying the natural logarithm and using logarithmic rules, we get: \[ n \times \ln(0.75) < \ln(0.01) \] Since \[ \ln(0.75) \] is negative, dividing by it reverses the inequality sign: \[ n > \frac{\ln(0.01)}{\ln(0.75)} \] When computed, \[ n \] turns out to be approximately 17 years. Therefore, it takes around 17 years for the pollutant level to reduce below 1% of its original amount.
Understanding these steps ensures you can solve similar problems involving exponential decay and inequalities.

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Most popular questions from this chapter

Find \(C\) and \(a\) such that the function \(f(x)=C a^{x}\) satisfies the given conditions. a. \(f(0)=6\) and for each unit increase in \(x,\) the output is multiplied by 1.2 . b. \(f(0)=10\) and for each unit increase in \(x,\) the output is multiplied by 2.5

(Graphing program recommended.) According to Mexico's National Institute of Geography, Information and Statistics, in 2005 in Mexico's Quintana Roo state, where the tourist industry in Cancun has created a boom economy, the population was about 1.14 million and growing at a rate of \(5.3 \%\) per year. In 2005 in Mexico's Baja California state, where many labor-intensive industries are located next to the California border, the population was about 2.8 million and growing at a rate of \(3.3 \%\) per year. The nation's capital, Mexico City, had 19.2 million inhabitants as of 2005 and the population was declining at a rate of \(0.012 \%\) per year. a. Construct three exponential functions to model the growth or decay of Quintana Roo, Baja California, and Mexico City. Identify your variables and their units. b. Use your functions to predict the populations of Quintana Roo, Baja Califomia, and Mexico City in \(2010 .\)

Determine which of the following functions are exponential. Identify each exponential function as representing growth or decay and find the vertical intercept. a. \(A=100\left(1.02^{t}\right)\) b. \(f(x)=4\left(3^{x}\right)\) c. \(g(x)=0.3\left(10^{x}\right)\) d. \(y=100 x+3\) e. \(M=2^{p}\) f. \(y=x^{2}\)

The future value \(V\) of a savings plan, where regular payments \(P\) are made \(n\) times to an account in which the interest rate, \(i,\) is compounded each payment period, can be calculated using the formula $$ V=P \cdot \frac{(1+i)^{n}-1}{i} $$ The total number of payments, \(n,\) equals the number of payments per year, \(m,\) times the number of years, \(t,\) so $$ n=m \cdot t $$ The interest rate per compounding period, \(i\), equals the annual interest rate, \(r,\) divided by the number of compounding periods a year, \(m,\) so $$ i=r / m $$ a. Substitute \(n=m \cdot t\) and \(i=r / m\) in the formula for \(V\), getting an expression for \(V\) in terms of \(m, t,\) and \(r\). b. If a parent plans to build a college fund by putting \(\$ 50\) a month into an account compounded monthly with a \(4 \%\) annual interest rate, what will be the value of the account in 17 years? c. Solve the original formula for \(P\) as a function of \(V, i,\) and \(n\). d. Now you are able to find how much must be paid in every month to meet a particular final goal. If you estimate the child will need \(\$ 100,000\) for college, what monthly payment must the parent make if the interest rate is the same as in part (b)?

(Graphing program recommended.) Below is a table of values for \(y=500(3)^{x}\) and for \(\log y\) $$ \begin{array}{rrr} \hline x & y & \log y \\ \hline 0 & 500 & 2.699 \\ 5 & 121,500 & 5.085 \\ 10 & 29,524,500 & 7.470 \\ 15 & 7.17 \cdot 10^{9} & 9.856 \\ 20 & 1.74 \cdot 10^{12} & 12.241 \\ 25 & 4.24 \cdot 10^{14} & 14.627 \\ 30 & 1.03 \cdot 10^{17} & 17.013 \\ \hline \end{array} $$ a. Plot \(y\) vs. \(x\) on a linear scale. Remember to identify the largest number you will need to plot before setting up axis scales. b. Plot \(\log y\) vs. \(x\) on a semi-log plot with a log scale on the vertical axis and a linear scale on the horizontal axis. c. Rewrite the \(y\) -values as powers of \(10 .\) How do these values relate to \(\log y ?\)
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