/*! 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 11 For Problems \(1-16,\) solve for... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For Problems \(1-16,\) solve for \(t\) using natural logarithms. $$40=100 e^{-0.03 t}$$

Short Answer

Expert verified
The solution is approximately \( t = 30.66 \).

Step by step solution

01

Isolate the Exponential Term

In the equation \( 40 = 100 e^{-0.03 t} \), we start by isolating the exponential term involving \( t \). Divide both sides by 100 to get \( 0.4 = e^{-0.03 t} \).
02

Apply Natural Logarithm

Take the natural logarithm of both sides to undo the exponential. This gives us \( \ln(0.4) = \ln(e^{-0.03 t}) \).
03

Simplify the Equation

Using the property \( \ln(e^x) = x \), simplify the right side to get \( \ln(0.4) = -0.03 t \).
04

Solve for t

To find \( t \), divide both sides by \(-0.03\). Thus, \( t = \frac{\ln(0.4)}{-0.03} \).
05

Calculate the Value

Now compute \( t \) using a calculator. Calculate \( t = \frac{\ln(0.4)}{-0.03} \), which gives \( t \approx 30.66 \).

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.

Exponential Equations
Exponential equations are mathematical expressions where a variable appears in the exponent. These equations often describe real-world situations such as population growth, radioactive decay, or cooling processes. In these equations, the base is usually a constant and the unknown variable is in the power. For example, in the equation \( 40 = 100 e^{-0.03 t} \), the variable \( t \) appears in the exponent and the base is the constant \( e \).
Solving exponential equations typically involves transforming them into a form where we can apply logarithms. The goal is to isolate the term with the exponent, making it easier to work with logarithmic identities. This process often involves dividing or multiplying both sides of the equation to simplify, as seen in the step where we divided by 100 to isolate the exponential term \( e^{-0.03 t} \).
Recognizing exponential equations in various forms is key. They might not always appear straightforwardly but understanding they involve powers with an unknown exponent helps in recognizing how to approach solving them.
Solving Equations
Solving equations is about finding the value of unknown variables that make an equation true. With exponential equations, this process can be a bit more complex since the unknown is in the exponent.
To solve an equation like \( 0.4 = e^{-0.03 t} \), start by simplifying the equation so that the term involving the unknown is isolated. Once isolated, logarithms can be applied to both sides, effectively helping to 'bring down' the exponent, turning it into a more manageable form.
The application of the natural logarithm, \( \, \, \ln \), is a strategic choice for equations involving the base \( e \), because of logarithmic identities. This process converts the exponent into a linear form, helping transform the equation into one that can easily be solved through basic algebraic steps, like division, as performed to solve for \( t \) in the exercise.
Using Logarithms
Logarithms are the inverse operations of exponentiation, which means they can "undo" exponential functions. Natural logarithms, denoted as \( \ln \), are particularly useful when dealing with exponential equations with base \( e \).
When you take the natural logarithm of both sides of an equation like \( e^{-0.03 t} \), you're using the identity \( \ln(e^x) = x \) to simplify the process. This turns the exponential expression into a linear one, greatly simplifying the task of solving for the variable in the exponent.
After applying the natural logarithm and using properties to simplify, basic algebra can be employed to solve for the unknown variable. This is seen in our solution where \( \ln(0.4) = -0.03 t \) quickly turns into a simple division problem, making it much more straightforward to find the actual value of \( t \). Logarithms thus serve as a powerful tool in manipulating and solving exponential equations.

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

Write the functions in Problems \(21-24\) in the form \(P=P_{0} a^{t}\) Which represent exponential growth and which represent exponential decay? $$P=P_{0} e^{0.2 t}$$

The Hershey Company is the largest US producer of chocolate. In \(2011,\) annual net sales were 6.1 billion dollars and were increasing at a continuous rate of \(4.2 \%\) per year. \(^{65}\) (a) Write a formula for annual net sales, \(S,\) as a function of time, \(t,\) in years since 2011 (b) Estimate annual net sales in 2015 (c) Use a graph to estimate the year in which annual net sales are expected to pass 8 billion dollars and check your estimate using logarithms.

In Problems \(29-30,\) a quantity \(P\) is an exponential function of time \(t .\) Use the given information about the function \(P=P_{0} e^{k t}\) to: (a) Find values for the parameters \(k\) and \(P_{0}\). (b) State the initial quantity and the continuous percent rate of growth or decay. \(P=140\) when \(t=3\) and \(P=100\) when \(t=1\)

When the price, \(p,\) charged for a boat tour was $$ 25,\( the average number of passengers per week, \)N,\( was $$500 . When the price was reduced to $$ 20,\) the average number of passengers per week increased to \(650 .\) Find a formula for the demand curve, assuming that it is linear.

For children and adults with diseases such as asthma, the number of respiratory deaths per year increases by \(0.33 \%\) when pollution particles increase by a microgram per cubic meter of \operatorname{air}^{66}. (a) Write a formula for the number of respiratory deaths per year as a function of quantity of pollution in the air. (Let \(Q_{0}\) be the number of deaths per year with no pollution.) (b) What quantity of air pollution results in twice as many respiratory deaths per year as there would be without pollution?

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.