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Chapter 5: Problem 27

Find all the real-number roots of each equation. In each case, give an exact expression for the root and also (where appropriate) a calculator approximation rounded to three decimal places. \(e^{x}-e^{-x}=1 \quad\) Hint: Multiply both side by \(e^{x}.\)

Short Answer

Expert verified
The exact root is \(\ln\left(\frac{1 + \sqrt{5}}{2}\right)\) and the approximation is 0.481.

Step by step solution

01

Multiply Through by \(e^x\)

Start with the equation \(e^x - e^{-x} = 1\). To eliminate the term \(e^{-x}\), multiply both sides by \(e^x\). This yields \((e^x)^2 - 1 = e^x\).
02

Rearrange the Equation

Rearrange the equation \((e^x)^2 - 1 = e^x\) as \((e^x)^2 - e^x - 1 = 0\), which is a quadratic equation in terms of \(e^x\).
03

Use the Quadratic Formula

Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve \(t^2 - t - 1 = 0\) by letting \(t = e^x\), \(a = 1\), \(b = -1\), and \(c = -1\).
04

Apply the Quadratic Formula

Calculate \(t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-1)}}{2 \times 1}\) which simplifies to \(t = \frac{1 \pm \sqrt{5}}{2}\). This gives two possible values for \(e^x\): \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\).
05

Determine Valid Solution for \(e^x\)

Since \(e^x > 0\), only \(\frac{1 + \sqrt{5}}{2}\) is valid. The other value \(\frac{1 - \sqrt{5}}{2}\) is negative and cannot be \(e^x\).
06

Solve for \(x\)

Take the natural logarithm of both sides to solve for \(x\):\[ x = \ln\left(\frac{1 + \sqrt{5}}{2}\right) \].
07

Calculate and Approximate the Root

Use a calculator to find the numerical approximation of \(x\): \[ x \approx \ln(1.6180339887) \approx 0.481 \] (rounded to three decimal places).

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Key Concepts

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

Real-Number Roots
When solving exponential equations like the one we have, understanding real-number roots becomes crucial. Real roots are simply solutions to the equation where the solution is a real number, as opposed to an imaginary or complex number. In this context, determining whether a solution is a real number involves examining the possible outcomes of its expression.
  • In our specific exercise, we used an equation involving exponential forms, namely, \(e^x\) and \(e^{-x}\).
  • This was transformed into a quadratic equation, which helps to identify real-number roots more easily.
  • By applying techniques such as the quadratic formula, we narrowed down the values that \(e^x\) could take, emphasizing that \(e^x\) must always be positive due to its nature as an exponential term.
Recognizing valid real-number roots is essential, as not all solutions from a quadratic form can be interpreted as such, because negative square roots are not possible within this scenario.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). In our scenario, the quadratic formula was used after transforming the exponential equation into a quadratic form:
  • The formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a\), \(b\), and \(c\) are coefficients in the quadratic equation \(at^2 + bt + c = 0\).
  • In the example we considered, \(a = 1\), \(b = -1\), and \(c = -1\). Using the formula helped to find potential solutions for \(e^x\).
  • The result was two values: \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\).
Applying the quadratic formula requires a clear understanding of terms and solving under given conditions, such as checking for real roots when dealing with exponential terms. The roots found represent potential values for the transformed variable, which, in this exercise, was \(e^x\).
Natural Logarithm
The natural logarithm, represented by \(\ln\), is essential when solving exponential equations for solving back to \(x\).
  • In this exercise, after finding that \(e^x = \frac{1 + \sqrt{5}}{2}\), we needed to solve for \(x\).
  • This was done by taking the natural logarithm (log base \(e\)) of both sides: \[x = \ln\left(\frac{1 + \sqrt{5}}{2}\right)\].
  • The natural logarithm helps to simplify expressions that include exponential terms, transforming them so their power can be solved directly.
  • Finally, we used a calculator to determine the decimal approximation of the natural logarithm, yielding \(x \approx 0.481\), rounded to three decimal places.
The application of the natural logarithm bridges the gap between exponential expressions and their solutions in terms of \(x\), making it an essential operation in exponential equations.

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Most popular questions from this chapter

Solve each equation. $$3(\ln x)^{2}-\ln \left(x^{2}\right)-8=0$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$4\left(10-e^{x}\right) \leq-3$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$6\left(5-1.6^{x}\right) \geq 13$$

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

If \(a^{2}+b^{2}=7 a b,\) where \(a\) and \(b\) are positive, show that $$\log \left[\frac{1}{3}(a+b)\right]=\frac{1}{2}(\log a+\log b)$$ no matter which base is used for the logarithms (but it understood that the same base is used throughout).
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