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Chapter 3: Problem 64

Solve each equation for \(x\) $$ \text { (a) }\ln \left(x^{2}-1\right)=3 \quad \text { (b) } e^{2 x}-3 e^{x}+2=0 $$

Short Answer

Expert verified
(a) \(x = \pm \sqrt{e^3 + 1}\); (b) \(x = 0\) or \(x = \ln(2)\).

Step by step solution

01

Solve for x in Part (a)

Given the equation \(\ln(x^2 - 1) = 3\). Exponentiate both sides to remove the natural logarithm: \(e^{\ln(x^2 - 1)} = e^3\). This gives \(x^2 - 1 = e^3\).
02

Isolate \(x^2\) in Part (a)

Add 1 to both sides of the equation: \(x^2 = e^3 + 1\).
03

Solve for x in Part (a)

Take the square root of both sides: \(x = \pm \sqrt{e^3 + 1}\). Therefore, \( x = \pm \sqrt{e^3 + 1} \).
04

Solve for e^x in Part (b) - Substitution

Given the equation \(e^{2x} - 3e^{x} + 2 = 0\), set \(y = e^{x}\) so that the equation becomes \(y^2 - 3y + 2 = 0\).
05

Factor the Quadratic Equation in Part (b)

Factor the quadratic to solve for \(y\): \((y - 1)(y - 2) = 0\). This means \(y = 1\) or \(y = 2\).
06

Solve for x in Part (b)

Replace \(y\) with \(e^x\) to get \(e^x = 1\) or \(e^x = 2\). Solve for \(x\):- For \(e^x = 1\), \(x = \ln(1) = 0\).- For \(e^x = 2\), \(x = \ln(2)\). Therefore, \(x = 0\) or \(x = \ln(2)\).

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Key Concepts

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

Natural Logarithm
The natural logarithm, often represented as \( \ln \), is a logarithm with a base of Euler's number \( e \), which is approximately 2.71828. This concept is commonly used to solve exponential equations and appears frequently in calculus and physics.
  • The natural logarithm of a number is the power to which \( e \) must be raised to obtain that number. For example, if \( \ln(a) = b \), then \( e^b = a \).
  • Logarithmic functions are the inverse of exponential functions. This means \( \ln(e^x) = x \) for all \( x \) and \( e^{\ln x} = x \), assuming \( x > 0 \).
In the given exercise, you had to solve \( \ln(x^2 - 1) = 3 \). To do this, you convert the logarithmic equation into its exponential form: \( x^2 - 1 = e^3 \). By recognizing this relationship, you can effectively deal with various problems involving natural logarithms by translating them into exponential equations.
Exponential Equations
Exponential equations are mathematical expressions where the variable is in the exponent. They form the basis of many real-world situations, such as compound interest, population growth, and radioactive decay. Understanding the properties of exponential functions helps in solving these equations effectively.
  • Exponential growth and decay describe processes that increase or decrease at rates proportional to their current value. Common examples are \( a^x \), where \( a \) is a constant base.
  • The most natural base for exponential functions is Euler's number \( e \), due to its unique properties in calculus.
To solve an equation like \( e^{2x} - 3e^x + 2 = 0 \), it's useful to substitute \( e^x \) with a simpler variable, such as \( y \). This transforms the equation into a more manageable quadratic form: \( y^2 - 3y + 2 = 0 \). Making this substitution can simplify the problem-solving process and highlight the relationship between exponential and quadratic equations.
Quadratic Equations
Quadratic equations are polynomial equations of the second degree, typically in the form \( ax^2 + bx + c = 0 \). These equations are fundamental in algebra and arise naturally in various fields, including physics, engineering, and economics.
  • A quadratic equation can have two solutions, one solution, or no real solutions, depending on the discriminant, \( b^2 - 4ac \).
  • Factoring is a common method to solve quadratic equations, especially when the equation can be expressed as the product of two binomials, like \((y - 1)(y - 2) = 0\) in the exercise.
  • In some cases, quadratic equations can be solved using the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \).
When solving an exponential equation with quadratic transformations, like substituting \( e^x = y \), you effectively reduce the problem to solving the quadratic equation \( y^2 - 3y + 2 = 0 \). This helps break down complex problems into simpler, more familiar terms that are easier to solve.

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Most popular questions from this chapter

If \(x=\ln (\sec \theta+\tan \theta),\) show that \(\sec \theta=\cosh x\)

Differentiate the function. $$ y=10^{1-x^{2}} $$

Differentiate the function. $$ y=x^{2} e^{-1 / x} $$

(a) What are the domain and range of \(f ?\) (b) What is the \(x\) -intercept of the graph of \(f ?\) (c) Sketch the graph of \(f .\) $$ f(x)=\ln (x-1)-1 $$

$$f(x)=\left\\{\begin{array}{ll}{|x|^{x}} & {\text { if } x \neq 0} \\ {1} & {\text { if } x=0}\end{array}\right.$$ (a) Show that \(f\) is continuous at 0 . (b) Investigate graphically whether \(f\) is differentiable at 0 by zooming in several times toward the point \((0,1)\) on the graph of \(f .\) (c) Show that \(f\) is not differentiable at \(0 .\) How can you reconcile this fact with the appearance of the graphs in part (b)?
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