/*! 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 29 Solve the exponential equation. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the exponential equation. Round to three decimal places, when needed. $$9-e^{x^{2}-1}=2$$

Short Answer

Expert verified
The solutions to the equation are \(x = 2.807\) and \(x = -2.807\), rounded to three decimal places.

Step by step solution

01

Isolate the Exponential Function

Rearrange the given equation \(9-e^{x^{2}-1}=2\) to isolate \(e^{x^{2}-1}\) on one side of the equation. Subtract 2 from both sides: \(9-2 = e^{x^{2}-1} \Rightarrow e^{x^{2}-1}=7\)
02

Take Natural Logarithm on Both Sides

To get rid of the exponential function and to make \(x\) manageable, take the natural logarithm on both sides of the equation: \(\ln(e^{x^{2}-1}) = \ln(7)\)
03

Simplify using Properties of Logarithms

Using the property that \(\ln(e^{a}) = a\), simplify the left side of the equation. Doing so, we get \(x^{2}-1 = \ln(7)\)
04

Solve for \(x\)

Rearrange the above equation to solve for \(x\). Add 1 to both sides, and then take the square root: \(x^{2} = \ln(7) + 1\), then \(x = \pm\sqrt{\ln(7)+1}\).
05

Round To Three Decimal Places

After solving for \(x\), round the values to three decimal places as required by the question. Also note that we have used \(\pm\) while taking the square root, hence, there will be two solutions for \(x\): \(x = \sqrt{\ln(7)+1}\) and \(x = -\sqrt{\ln(7)+1}\). This gives \(x \approx \pm 2.807\) (rounded to three decimal places)

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.

Natural Logarithm
The natural logarithm is denoted as \( \ln \) and has a base of \( e \), which is an irrational and transcendental number approximately equal to 2.71828. It is particularly useful in solving exponential equations where the exponential term has \( e \) as the base.

In our exercise, after isolating the term \( e^{x^{2}-1} \), taking the natural logarithm on both sides becomes a key step. This is because the natural logarithm function, \( \ln \), is the inverse of the exponential function \( e^{x} \).

By applying \( \ln \) to \( e^{x^{2}-1} = 7 \), we get \( \ln(e^{x^{2}-1}) = \ln(7) \). This effectively "cancels out" the \( e \) on the left side, leaving us with \( x^{2} - 1 \), and thus simplifying the problem significantly.
Properties of Logarithms
Logarithms come with various properties that simplify complex calculations. One of the most beneficial properties is the power rule: \( \ln(e^{a}) = a \).

This property is pivotal in our step-by-step solution where \( \ln(e^{x^{2}-1}) = \ln(7) \). It simplifies to \( x^{2} - 1 = \ln(7) \), allowing us to easily handle the expression involving \( x \).
  • **Product Rule**: \( \ln(AB) = \ln(A) + \ln(B) \)
  • **Quotient Rule**: \( \ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B) \)
  • **Power Rule**: \( \ln(A^{b}) = b \cdot \ln(A) \)
These rules are powerful tools in algebra and calculus. They are important for transforming and simplifying equations that involve logarithms.
Rounding Numbers
Rounding numbers is a mathematical method used to simplify figures to make them easier to work with. It is especially important in ensuring that results reflect the precision required by a particular problem.

When rounding to three decimal places, we follow these general steps:
  • Identify the digit at the third decimal place.
  • Look at the digit immediately to the right; if it is 5 or greater, increase the third decimal place by 1.
  • If it is less than 5, keep the third decimal place unchanged.
In our exercise, after determining the solutions for \( x \), we rounded the values to get \( x \approx \pm 2.807 \), providing clarity while adhering to the problem's instructions.

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

Applications In this set of exercises, you will use inverse functions to study real-world problems. In economics, the demand function gives the price \(p\) as a function of the quantity \(q .\) One example of a demand function is \(p=100-0.1 q .\) However, mathematicians tend to think of the price as the input variable and the quantity as the output variable. How can you take this example of a demand function and express \(q\) as a function of p?

Do all linear functions have inverses? Explain.

If the graph of a function \(f\) is symmetric with respect to the \(y\) -axis, can \(f\) be one-to-one? Explain.

Refer to the following. The pH of a solution is defined as \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] .\) The concentration of hydrogen ions, \(\left[\mathrm{H}^{+}\right]\), is given in moles per liter, where one mole is equal to \(6.02 \times 10^{23}\) molecules. What is the concentration of hydrogen ions in a solution that has a pH of \(1.5 ?\)

A new car that costs $$\$ 25,000$$ depreciates to $$80 \%$$ of its value in 3 years. (a) Assume the depreciation is linear. What is the linear function that models the value of this car \(t\) years after purchase? (b) Assume the value of the car is given by an exponential function \(y=A e^{h t},\) where \(A\) is the initial price of the car. Find the value of the constant \(k\) and the exponential function. (c) Using the linear model found in part (a), find the value of the car 5 years after purchase. Do the same using the exponential model found in part (b). (d) Graph both models using a graphing utility. Which model do you think is more realistic, and why?

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.