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Chapter 1: Problem 1

Determine the doubling times of the following exponential equations. (a) \(y=2^{t}\) (b) \(y=2^{3 t}\) (c) \(y=2^{0.1 t}\) (d) \(y=10^{t}\) (e) \(y=10^{3 t}\) (f) \(y=10^{0.1 t}\)

Short Answer

Expert verified
(a) 1, (b) 1/3, (c) 10, (d) ~0.301, (e) ~0.1003, (f) ~3.01

Step by step solution

01

Understanding Doubling Time

The doubling time of an exponential function is the time it takes for its value to double. For an equation of the form \(y = a^{bt}\), the doubling time \(T_d\) can be found using the formula \(a^{bT_d} = 2\). Solving this lets us determine how long it takes for the function to double.
02

Equation (a): Simplifying and Solving

For equation (a) \(y = 2^t\), we set up the equation \(2^{T_d} = 2\). Solve for \(T_d\) by comparing the exponents on both sides, which gives \(T_d = 1\).
03

Equation (b): Simplifying and Solving

For equation (b) \(y = 2^{3t}\), set the equation \(2^{3T_d} = 2\) to find \(T_d\). Dividing exponents, we get \(3T_d = 1\), hence \(T_d = \frac{1}{3}\).
04

Equation (c): Simplifying and Solving

For equation (c) \(y = 2^{0.1t}\), set \(2^{0.1T_d} = 2\). Solve for \(T_d\) by calculating \(0.1T_d = 1\), leading to \(T_d = 10\).
05

Equation (d): Simplifying and Solving

For equation (d) \(y = 10^t\), the equation is set as \(10^{T_d} = 2\). Taking the log (base 10) on both sides gives \(T_d = \log_{10}(2)\), approximately \(T_d \approx 0.301\).
06

Equation (e): Simplifying and Solving

For equation (e) \(y = 10^{3t}\), set \(10^{3T_d} = 2\). Solve via \(3T_d = \log_{10}(2)\), allowing us \(T_d = \frac{\log_{10}(2)}{3}\), which approximates to \(T_d \approx 0.1003\).
07

Equation (f): Simplifying and Solving

For equation (f) \(y = 10^{0.1t}\), set \(10^{0.1T_d} = 2\). This translates to \(0.1T_d = \log_{10}(2)\), resulting in \(T_d = \frac{\log_{10}(2)}{0.1}\), leading to \(T_d \approx 3.01\).

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Key Concepts

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

Exponential Function
An exponential function is a type of mathematical relation that describes situations where a quantity grows or decays at a constant rate. This is typically expressed as \( y = a^{bt} \), where \( a \) is the base, \( b \) is a constant, and \( t \) is the time variable. This type of function often models real-world scenarios like population growth, radioactive decay, or interest compounding in finance. Understanding exponential functions is crucial for solving problems related to growth and decay patterns, especially when needing to determine metrics like doubling time.
  • The base, \( a \), dictates the rate of growth or decay.
  • The exponent, in this case \( bt \), affects how fast the doubling or halving occurs.
  • Exponential functions are characterized by their rapid increase or decrease.
Equations
Equations are mathematical statements that assert the equality of two expressions. In the context of exponential functions, equations help us find the value of a variable, such as the time it takes for a function's value to double. Solving equations involving exponential functions often requires setting the function equal to a specific value and solving for the unknown.
For example, to find the doubling time in the form \( y = a^{bt} \), you would set \( a^{bT_d} = 2 \) where \( T_d \) is the doubling time. Solving this equation involves understanding the properties of the exponential function and might include using logarithms to isolate the exponent.
  • Equations bridge the initial setup of a problem with its solution.
  • Involves applying algebraic and sometimes numerical techniques to find unknowns.
  • Can configure exponential function equations to match practical scenarios like doubling time or half-life problems.
Logarithms
Logarithms are the inverse of exponential functions. They are used to solve equations involving exponentials, especially when isolating the variable in the exponent. If you have an equation of the form \( a^x = b \), taking the logarithm of both sides enables you to solve for \( x \).
In the doubling time problems seen in the exercise, logarithms help to handle equations like \( 10^{T_d} = 2 \), where taking the \( \log_{10} \) of both sides gives \( T_d = \log_{10}(2) \).
  • Using logarithms simplifies solving exponentials by reducing powers to multipliers.
  • They provide an essential tool for problems where exponential terms need to be isolated.
  • Logarithms help convert multiplication into addition, which is simpler to solve.
Calculus
Calculus is a branch of mathematics that studies change. While calculus itself may not always be necessary to solve basic exponential doubling time problems, it provides a deeper understanding of how exponential functions behave. For example, understanding derivatives can help you analyze how fast a function changes over time.
Though the exercise primarily requires algebra and logarithms, calculus equips us with tools such as differentiation and integration that allow for in-depth analysis of growth rates and area under curves of exponential functions.
  • Calculus helps in understanding the rate at which changes occur in the exponentials.
  • Differentiation can provide insights into the speed of growth or decay of functions.
  • Integration allows for comprehensive analysis over continuous intervals, including predictive modeling.

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Most popular questions from this chapter

The polymerase chain reaction is a means of making multiple copies of a DNA segment from only a minute amounts of original DNA. The procedure consists of a sequence of, say, 30 cycles in which each segment present at the beginning of a cycle is duplicated once; at the end of the cycle that segment and one copy is present. Introduce notation and write a difference equation with initial condition from which the amount of DNA present at the end of each cycle can be computed. Suppose you begin with 1 picogram \(=0.000000000001 \mathrm{~g}\) of DNA. How many grams of DNA would be present after 30 cycles.

A one-liter flask contains one liter of distilled water and \(2 \mathrm{~g}\) of salt. Repeatedly, 50 \(\mathrm{ml}\) of solution are removed from the flask and discarded after which \(50 \mathrm{ml}\) of distilled water are added to the flask. Introduce notation and write a dynamic equation that will describe the change of salt in the beaker each cycle of removal and replacement. How much salt is in the beaker after 20 cycles of removal?

Two kilos of a fish poison, rotenone, are mixed into a lake which has a volume of \(100 \times 20 \times 2=4000\) cubic meters. No water flows into or out of the lake. Fifteen percent of the rotenone decomposes each day. a. Write a mathematical model that describes the daily change in the amount of rotenone in the lake. b. Let \(R_{0}, P_{1}, R_{2}, \cdots\) denote the amounts of rotenone in the lake, \(P_{t}\) being the amount of poison in the lake at the beginning of the \(t \underline{h}\) day after the rotenone is administered. Write a dynamic equation representative of the mathematical model. c. What is \(R_{0} ?\) Compute \(R_{1}\) from your dynamic equation. Compute \(R_{2}\) from your dynamic equation. d. Find a solution equation for your dynamic equation.

Equation 1.30 $$ P_{t+1}-P_{t}=b $$ represents a large number of equations $$ \begin{array}{r} P_{1}-P_{0}=b \\ P_{2}-P_{1}=b \\ P_{3}-P_{2}=b \\ \vdots \quad \vdots \quad \vdots \quad \vdots \\ P_{n-1}-P_{n-2}=b \\ P_{n}-P_{n-1}=b \end{array} $$ Add these equations to obtain $$ P_{n}=P_{0}+n b $$ Substitute \(t\) for \(n\) to obtain $$ P_{t}=P_{0}+t b $$

Show that \(y=A B^{t}\) with \(B<1\) has a half-life of $$ T_{\text {Half }}=\frac{\log \frac{1}{2}}{\log B}=\frac{-\log 2}{\log B} $$
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