/*! 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 77 Use a graphing utility to graph ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$e^{0.09 t}=3$$

Short Answer

Expert verified
The graphical solution would be around \( t \approx 10.416 \). The confirmed algebraic solution is \( t = \frac{ln(3)}{0.09} \approx 10.416 \). Verification is done by substitifying the obtained value back into the original equation and making sure that it holds.

Step by step solution

01

Graph the Function

Use a graphing utility to graph the equation \( e^{0.09 t} = 3 \). The graph of this equation would show where the function intersects the line \( y = 3 \), the points of intersection represent the solution of the equation.
02

Approximate the Solution

By looking at the graph, observe the point(s) at which the function crosses the line \( y = 3 \). Approximate the value of \( t \) at this point up to three decimal places.
03

Algebraically Solve the Equation

To verify the solution, solve the equation algebraically. To do so, apply the natural logarithm (ln) to both sides of the equation, resulting in \( 0.09 t = ln(3) \). Then, solve for \( t \) by dividing both sides by 0.09, and compare this result with the approximation obtained from the graph.

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!

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

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln (x+1)-\ln (x-2)=\ln x$$

Use the following information for determining sound intensity. The level of sound \(\boldsymbol{\beta}\), in decibels, with an intensity of \(I\), is given by \(\boldsymbol{\beta}=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is an intensity of \(10^{-12}\) watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. In Exercises 65 and 66 , find the level of sound \(\boldsymbol{\beta}\). (a) \(I=10^{-11}\) watt per \(\mathrm{m}^{2}\) (rustle of leaves) (b) \(I=10^{2}\) watt per \(\mathrm{m}^{2}\) (jet at 30 meters) (c) \(I=10^{-4}\) watt per \(\mathrm{m}^{2}\) (door slamming) (d) \(I=10^{-2}\) watt per \(\mathrm{m}^{2}\) (siren at 30 meters)

Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Compute \(\left[\mathrm{H}^{+}\right]\) for a solution in which \(\mathrm{pH}=3.2\).

The IQ scores for a sample of a class of returning adult students at a small northeastern college roughly follow the normal distribution \(y=0.0266 e^{-(x-100)^{2} / 450}, 70 \leq x \leq 115,\) where \(x\) is the IQ score. (a) Use a graphing utility to graph the function. (b) From the graph in part (a), estimate the average IQ score of an adult student.

Determine the time necessary for $$\$ 1000$$to double if it is invested at interest rate \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$r=6.5 \%$$

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.