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Chapter 5: Problem 25

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. (a) \(e^{2 x}+2 e^{x}+1=0\) (b) \(e^{2 x}-2 e^{x}+1=0\) (c) \(e^{2 x}-2 e^{x}-3=0\) (d) \(e^{2 x}-2 e^{x}-4=0\)

Short Answer

Expert verified
(a) No real roots. (b) \(x = 0\). (c) \(x = \ln(3) \approx 1.099\). (d) \(x = \ln(1 + \sqrt{5}) \approx 1.443\).

Step by step solution

01

Solve Equation (a)

For the equation \( e^{2x} + 2e^{x} + 1 = 0 \), substitute \( u = e^{x} \). This transforms the equation into \( u^2 + 2u + 1 = 0 \).This is a perfect square trinomial and can be factored as \((u + 1)^2 = 0\). Therefore, \( u + 1 = 0 \), which implies \( u = -1 \).Since \( u = e^{x} \) and \( e^{x} \) is always positive, there are no real solutions.
02

Solve Equation (b)

For the equation \( e^{2x} - 2e^{x} + 1 = 0 \), substitute \( u = e^{x} \) to get \( u^2 - 2u + 1 = 0 \).This is a perfect square trinomial and can be factored as \((u - 1)^2 = 0\). Thus, \( u - 1 = 0 \), and \( u = 1 \).Since \( u = e^{x} \), we have \( e^{x} = 1 \), giving us \( x = 0 \). The calculator approximation is also \( 0 \).
03

Solve Equation (c)

For the equation \( e^{2x} - 2e^{x} - 3 = 0 \), substitute \( u = e^{x} \) to get \( u^2 - 2u - 3 = 0 \).This can be factored into \((u - 3)(u + 1) = 0 \). Therefore, \( u = 3 \) or \( u = -1 \).Since \( u = e^{x} \) and \( e^{x} \) is always positive, only \( u = 3 \) is viable. Thus, \( e^{x} = 3 \), which implies \( x = \ln(3) \). Using a calculator, \( x \approx 1.099 \).
04

Solve Equation (d)

For the equation \( e^{2x} - 2e^{x} - 4 = 0 \), substitute \( u = e^{x} \) to get \( u^2 - 2u - 4 = 0 \).Apply the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -4 \):\[u = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 16}}{2}\]\[u = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}\]Since \( u = e^{x} > 0 \), we have \( e^{x} = 1 + \sqrt{5} \). Thus, \( x = \ln(1 + \sqrt{5}) \). Using a calculator, \( x \approx 1.443 \).

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Key Concepts

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

Quadratic Equations
Quadratic equations form a core part of algebra and are essential in solving various mathematical problems. A quadratic equation is generally expressed in the format \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants. The solutions to a quadratic equation are known as the roots and are calculated using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows us to find real-number solutions by determining how the graph of the quadratic interacts with the x-axis.
  • The Discriminant: The part under the square root, \( b^2 - 4ac \), is called the discriminant. It tells us about the nature of the roots.
    • If the discriminant is positive, there are two distinct real roots.
    • If it is zero, there is one real root (or a repeated root).
    • If negative, the roots are not real-number roots but complex numbers.
  • Factoring: Quadratic equations can sometimes be solved by factoring when they easily decompose into a perfect square trinomial or binomials.
These methods help in solving the quadratic equations presented in the original exercise.
Exponential Functions
Exponential functions are powerful mathematical expressions where a constant base, usually \( e \) (approximately 2.718), is raised to a variable exponent. The general form of an exponential function is:\[ f(x) = a \cdot e^{bx} \]where \( a \) is a constant, and \( b \) determines the growth or decay rate.In the context of solving equations like \( e^{2x} - 2e^{x} + 1 = 0 \), transforming exponential expressions into quadratic form can simplify the process. This method involves:
  • Substitution: Letting \( u = e^x \) to convert an exponential function into a more familiar quadratic form \( u^2 - 2u + 1 = 0 \).
  • Natural Logarithms: Once \( u \) is found, you can solve \( e^x = u \) by taking the natural logarithm, \( x = \ln(u) \), to find \( x \) values.
Exponential functions are essential in mathematical modeling, showcasing exponential growth or decay phenomena, such as population growth or radioactive decay.
Perfect Square Trinomial
A perfect square trinomial is a quadratic expression that can be expressed as the square of a binomial. It takes the form:\[ a^2 + 2ab + b^2 = (a+b)^2 \]This concept is crucial for simplifying quadratic equations efficiently.To recognize a perfect square trinomial:
  • Check if the first and last terms are perfect squares.
  • Ensure the middle term is twice the product of the square roots of the first and last terms.For example, \( x^2 + 6x + 9 = (x+3)^2 \).
In the original exercises:
  • Equation (a) \( e^{2x} + 2e^x + 1 = (e^x+1)^2 = 0 \): This is a perfect square trinomial since it factors perfectly into a binomial square.
  • Equation (b) \( e^{2x} - 2e^x + 1 = (e^x-1)^2 = 0 \): Another example where recognizing the perfect square trinomial facilitates solving.
Perfect square trinomials ease the process of factoring, simplifying the step-by-step solution.
Factoring
Factoring is a technique used to break down algebraic expressions into simpler components called factors. It is particularly handy when solving quadratic equations, enabling straightforward solutions.The factoring process involves finding two binomials when multiplied together, recreate the original quadratic equation. For instance:
  • Determine if there is a common factor in the entire equation; factor that out first.
  • In a quadratic form \( ax^2 + bx + c \), find two numbers that multiply to \( ac \) (the product of the first and last coefficients) and add to \( b \) (the middle coefficient).
Factoring was effectively applied in the original step-by-step solutions:
  • For equation (c), the expression \( e^{2x} - 2e^x - 3 = 0 \) factors into \((e^x - 3)(e^x + 1) = 0\).
  • These factors provide potential solutions by setting each to zero and solving for \( x \), following the substitutions \( u = e^x \).
Understanding factoring enables the solving of quadratic equations by transforming them into manageable terms, advancing through substituting back to resolve for the original variables.

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Most popular questions from this chapter

In the text we showed that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is constant for all time intervals of unit length, \([t, t+1] .\) Recall that we did this by computing the relative change \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)\) and noting that the result was a constant, independent of \(t .\) (If you've completed the previous exercise, you've done this calculation for yourself.) Now consider a time interval of arbitrary length, \([t, t+d] .\) The relative change in the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) over this time interval is \([\mathcal{N}(t+d)-\mathcal{N}(t)] / \mathcal{N}(t) .\) Show that this quantity is a constant, independent of \(t .\) (The expression that you obtain for the constant will contain \(e\) and \(d\), but not \(t\) As a check on your work, replace \(d\) by 1 in the expression you obtain and make sure the result is the same as that in the text where we worked with intervals of length \(d=1 .)\)

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

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).

Use the half-life information to complete each table. (The formula \(\mathcal{N}=\mathcal{N}_{0} e^{k t}\) is not required.) (a) Polonium-210: half-life \(=138.4\) days $$\begin{array}{lllll}\hline t \text { (years) } & 0 & 138.4 & 276.8 & 415.2 \\\\\mathcal{N} \text { (grams) } & 0.4 & & & 0.025 \\\\\hline\end{array}$$ (b) Polonium-214: half-life \(=1.63 \times 10^{-4}\) second $$\begin{array}{lllll}\hline t \text { (seconds) } & 0 & & & 6.52 \times 10^{-4} \\\\\mathcal{N} \text { (grams) } & 0.1 & 0.05 & 0.025 & 0.0125 & \\\\\hline\end{array}$$

Use the following information on \(p H\) Chemists define pH by the formula pH \(=-\log _{10}\left[\mathrm{H}^{+}\right],\) where [H \(^{+}\) ] is the hydrogen ion concentration measured in moles per liter. For example, if \(\left[\mathrm{H}^{+}\right]=10^{-5},\) then \(p H=5 .\) Solutions with \(a\) pH of 7 are said to be neutral; a p \(H\) below 7 indicates an acid: and a pH above 7 indicates a base. (A calculator is helpful for Exercises 49 and 50.1 An unknown substance has a hydrogen ion concentration of \(3.5 \times 10^{-9} .\) Classify the substance as acid or base.
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