/*! 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 120 Find the approximate solution to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 120

Find the approximate solution to each equation. Round to four decimal places. $$\frac{1}{e^{x-1}}=5$$

Short Answer

Expert verified
-0.6094

Step by step solution

01

Rewrite the Equation

Start by rewriting the given equation \(\frac{1}{e^{x-1}}=5\) in a form that can be easily solved. Multiply both sides by \(e^{x-1}\) to isolate it: \(1=5e^{x-1}\).
02

Isolate the Exponential Term

Divide both sides of the equation by 5 to get: \(\frac{1}{5}=e^{x-1}\).
03

Apply the Natural Logarithm

Take the natural logarithm (\text{ln}) of both sides to solve for \(x\): \(\text{ln}(\frac{1}{5})=\text{ln}(e^{x-1})\). The right side simplifies to \(x-1\) because \(\text{ln}(e^y) = y\). So the equation becomes: \(\text{ln}(\frac{1}{5})=x-1\).
04

Solve for \(x\)

Solve for \(x\) by adding 1 to both sides: \(x = \text{ln}(\frac{1}{5}) + 1\).
05

Calculate the Value

Use a calculator to find \(\text{ln}(\frac{1}{5})\): \(\text{ln}(0.2) \approx -1.6094\). Then add 1: \(-1.6094 + 1 = -0.6094\).

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 91影视!

Key Concepts

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

Natural logarithms
The natural logarithm, denoted as \(\text{ln}\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Natural logarithms are widely used in mathematics, especially in solving equations involving exponential functions.
They are particularly useful because of the relationship \(\text{ln}(e^y) = y\), which simplifies many problems significantly.
In our example, we transformed the equation \( e^{x-1} = \frac{1}{5} \) into \(\text{ln}(e^{x-1}) = \text{ln}(\frac{1}{5})\). This simplifies quickly to \(x-1\) due to the logging property, making it easier to isolate and solve for \(x\).
Exponential functions
Exponential functions are mathematical functions of the form \(f(x) = a \times e^{bx}\), where \(e\) is the base of the natural logarithms, and \(a\) and \(b\) are constants.
They grow (or decay, if \(b < 0\)) very rapidly and are useful in modeling a variety of real-world phenomena, such as population growth, radioactive decay, and interest calculations.
In our exercise, we dealt with the exponential function \(e^{x-1}\). It complicated the equation initially, but we managed to rewrite it as \(\frac{1}{5} = e^{x-1}\), eventually transforming it to a simpler form understandable through logarithms.
Equation solving steps
Solving exponential equations often involves several clear steps. Let's recap the ones used in the exercise:
  • First, rewrite the equation to isolate the exponential term.
  • In our example, multiply both sides of \(\frac{1}{e^{x-1}} = 5\) by \(e^{x-1}\) to get \(1 = 5 e^{x-1}\).
  • Next, divide both sides by the constant (here, 5): \(\frac{1}{5} = e^{x-1}\).
  • Apply the natural logarithm to both sides to deal with the exponential term: \(\text{ln}(\frac{1}{5}) = x-1\).
  • Isolate \(x\) by performing the necessary arithmetic operations. For this exercise, simply add 1 to both sides to achieve \(x = \text{ln}(\frac{1}{5}) + 1\).
  • Finally, perform the calculations: \(\text{ln}(0.2) \approx -1.6094\), yielding a conclusion of \(-1.6094 + 1 = -0.6094\).
By systematically following these steps, you can solve many similar 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

To evaluate an exponential or logarithmic function we simply press a button on a calculator. But what does the calculator do to find the answer? The next exercises show formulas from calculus that are used to evaluate \(e^{x}\) and \(\ln (1+x)\). Infinite Series for Logarithms The following formula from calculus can be used to compute values of natural logarithms: $$\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots$$ where \(-1< x <1 .\) The more terms that we use from the formula, the closer we get to the true value of \(\ln (1+x)\) Find \(\ln (1.4)\) by using the first five terms of the series and compare your result to the calculator value for \(\ln (1.4)\)

Group Toss Have all students in your class stand and toss a coin. Those that get heads sit down. Those that are left standing toss again and all who obtain heads must sit down. Repeat until no one is left standing. For each toss record the number of coins that are tossed- for example, \((1,30),(2,14),\) and so on. Do not include a pair with zero coins tossed. Enter the data into a graphing calculator and use exponential regression to find an equation of the form \(y=a \cdot b^{x}\) that fits the data. What percent of those standing did you expect would sit down after each toss? Is the value of \(b\) close to this number?

Cooking a Roast James knows that to get well-done beef, it should be brought to a temperature of \(170^{\circ} \mathrm{F}\). He placed a sirloin tip roast with a temperature of \(35^{\circ} \mathrm{F}\) in an oven with a temperature of \(325^{\circ},\) and after 3 hr the temperature of the roast was \(140^{\circ} .\) How much longer must the roast be in the oven to get it well done? If the oven temperature is set at \(170^{\circ},\) how long will it take to get the roast well done? HINT The difference between the roast temperature and the oven temperature decreases exponentially.

Solve each problem. The population of the world doubled from 1950 to \(1987,\) going from 2.5 billion to 5 billion people. Using the exponential model, $$P=P_{0} e^{r t},$$ find the annual growth rate \(r\) for that period. Although the annual growth rate has declined slightly to \(1.63 \%\) annually, the population of the world is still growing at a tremendous rate. Using the initial population of 5 billion in 1987 and an annual rate of \(1.63 \%\), estimate the world population in the year 2010

A pond contains 2000 fish of which \(10 \%\) are bass. How many bass must be added so that \(20 \%\) of the fish in the pond are bass?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
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.