/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The percentage of smokers who re... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The percentage of smokers who receive telephone counseling and successfully quit smoking for \(t\) months is given by $$ P(t)=21.4(0.914)^{t} $$ Data: New England Journal of Medicine; California's Smoker's Hotline a) In what month will \(15 \%\) of those who quit and used telephone counseling still be smoke-free? b) In what month will \(5 \%\) of those who quit and used telephone counseling still be smoke-free?

Short Answer

Expert verified
For 15%, approximately 3.94 months. For 5%, approximately 16.18 months.

Step by step solution

01

Understand the Problem

We need to find the value of \( t \) (time in months) when the percentage of smokers remaining smoke-free is 15% and 5% using the function \( P(t)=21.4(0.914)^{t} \).
02

Set Up the Equation for 15%

To find when 15% of smokers are still smoke-free, set \( P(t) = 15 \). Thus, \( 15 = 21.4(0.914)^{t} \).
03

Solve for t (15%)

Divide both sides by 21.4:\( \frac{15}{21.4} = (0.914)^{t} \).Calculate \( \frac{15}{21.4} \) to get approximately 0.7019.Thus, \( 0.7019 = (0.914)^{t} \).
04

Apply the Logarithm (15%)

Take the natural logarithm of both sides:\( \ln(0.7019) = \ln((0.914)^{t}) \).Using the power rule of logarithms, \( \ln(0.7019) = t \ln(0.914) \).Thus, \( t = \frac{\ln(0.7019)}{\ln(0.914)} \).
05

Calculate t for 15%

Compute the natural logarithms:\( t = \frac{\ln(0.7019)}{\ln(0.914)} \approx \frac{-0.3541}{-0.0899} \approx 3.94 \).So, \( t \approx 3.94 \) months when 15% are smoke-free.
06

Set Up the Equation for 5%

To find when 5% of smokers are still smoke-free, set \( P(t) = 5 \). Thus, \( 5 = 21.4(0.914)^{t} \).
07

Solve for t (5%)

Divide both sides by 21.4:\( \frac{5}{21.4} = (0.914)^{t} \).Calculate \( \frac{5}{21.4} \) to get approximately 0.2336.Thus, \( 0.2336 = (0.914)^{t} \).
08

Apply the Logarithm (5%)

Take the natural logarithm of both sides:\( \ln(0.2336) = \ln((0.914)^{t}) \).Using the power rule of logarithms, \( \ln(0.2336) = t \ln(0.914) \).Thus, \( t = \frac{\ln(0.2336)}{\ln(0.914)} \).
09

Calculate t for 5%

Compute the natural logarithms:\( t = \frac{\ln(0.2336)}{\ln(0.914)} \approx \frac{-1.4541}{-0.0899} \approx 16.18 \).So, \( t \approx 16.18 \) months when 5% are smoke-free.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

logarithms
Logarithms are a powerful tool in mathematics, mostly used to solve exponential equations. A logarithm answers the question: 'To what exponent must a base be raised to produce a given number?' For example, the logarithm base 10 of 100 is 2, written as \(\text{log}_{10}(100) = 2\), because 10 raised to the power of 2 is 100. In natural logarithms, we use base e (approximately 2.718). The natural logarithm is denoted as \( \text{ln} \).
When dealing with exponential decay equations, logarithms help us isolate the variable in the exponent. For instance, to solve \((0.914)^{t}=0.7019\), we take the natural logarithm of both sides to obtain the equation \(\text{ln}(0.7019) = t \text{ln}(0.914)\). This simplification follows from the power rule, which states that \(\text{ln}(a^b) = b \text{ln}(a)\). By applying this rule, we can solve for the time variable \( t \). Understanding these properties of logarithms is key for solving exponential decay problems effectively.
exponential functions
Exponential functions are mathematical functions that grow or decay at a constant rate per unit interval. They can be expressed in the form \ (P(t) = P_0 a^t) \ where \( P_0 \) is the initial amount, \ (a) \ is the base representing the growth or decay factor, and \( t \) is the variable for time.
In the given exercise, the function \( P(t)=21.4(0.914)^t \) is used to model the exponential decay of smokers who remain smoke-free with telephone counseling. Here, \ (21.4) \ represents the initial percentage of smokers who quit, and \ (0.914) \ is the decay factor, indicating a decrease in the percentage of non-smokers over time.
Exponentials play a crucial role in various fields like biology, finance, and physics as they describe processes that evolve over time, either growing or decaying exponentially. The key property is that the rate of change is proportional to the current value, rendering the function consistent and predictable in behaviors modeled by it.
solving equations
Solving equations, particularly exponential equations, often involves several steps including manipulating the equation and using logarithms to isolate the variable. Here is some detailed advice on how to approach such problems:
  • Identify the structure of the equation. For exponential equations, look for forms like \ P(t) = P_0 a^t\.
  • Express the exponential part in a manner suitable for applying logarithms. For instance, converting \( a^t \) to \ \text{ln}(a^t) \ by taking the natural logarithm of both sides.
  • Use logarithmic properties to simplify the equation. For example, apply \ \text{ln}(a^b) = b \text{ln}(a) \ to bring the exponent down.
  • Isolate the variable, making sure to carefully perform algebraic operations on both sides.
In the provided exercise, this process was carried out efficiently. We took natural logarithms to solve for \( t \), transforming the exponential form into a linear equation for easier manipulation. Always verify your results by plugging back into the original equation to ensure consistency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The frequency, in hertz \((\mathrm{Hz}),\) of the \(n\) th key on an 88 -key piano is given by $$ f(n)=27.5(\sqrt[12]{2})^{n-1} $$ where \(n=1\) corresponds to the lowest key on the piano keyboard, an A. a) What number key on the keyboard has a frequency of \(440 \mathrm{Hz} ?\) b) How many keys does it take for the frequency to double?

Solve. If no solution exists, state this. $$ 3^{2 x}-3^{2 x-1}=18 $$

Solve \(\left|\log _{3} x\right|=2\)

Home Construction. The chemical urea formaldehyde was used in some insulation in houses built during the mid to late 1960 s. Unknown at the time was the fact that urea formaldehyde emitted toxic fumes as it decayed. The half-life of urea formaldehyde is 1 year. a) What is its decay rate? b) How long will it take \(95 \%\) of the urea formaldehyde present to decay?

Legend has it that because he objected to teenagers smoking, and because his first baseball card was issued in cigarette packs, the great shortstop Honus Wagner halted production of his card before many were produced. One of these cards was sold in 2008 for 1.62 million dollars. The same card was sold in 2013 for 2.1 million dollars. For the following questions, assume that the card's value increases exponentially, as it has for many years. a) Find the exponential growth rate \(k,\) and determine an exponential function that can be used to estimate the dollar value, \(V(t),\) of the card \(t\) years after 2008 b) Predict the value of the card in 2025 . c) What is the doubling time for the value of the card? d) In what year will the value of the card first exceed 5 million dollars?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.