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Chapter 4: Problem 21

Solve each exponential equation in Exercises \(1-26\) Express the solution set in terms of natural logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{2 x}-3 e^{x}+2=0$$

Short Answer

Expert verified
The solution to the equation \(e^{2x} - 3e^{x} + 2 = 0\) is \(x = ln(1)\) and \(x = ln(2)\), approximated as \(x \approx 0\) and \(x \approx 0.69\) respectively.

Step by step solution

01

Transform into a Quadratic Equation

The exponential equation \(e^{2x} - 3e^{x} + 2 = 0\) can be rewritten as a quadratic equation with the substitution \(e^{x} = y\). This yields a quadratic equation \(y^{2} - 3y + 2 = 0\).
02

Solve the Quadratic Equation

The quadratic equation can be solved using the quadratic formula \(y = ( -b \pm \sqrt{b^2-4ac} ) / 2a\), where a, b, and c are coefficients from our equation \(y^{2} - 3y + 2 = 0\). So, \(a=1\), \(b=-3\), and \(c=2\). This results in solutions \(y_1 = 1\) and \(y_2 = 2\).
03

Back Substitute

Since we made the substitution \(e^{x}=y\) we should now back substitute. The result yields two equations: \(e^{x}=1\) and \(e^{x}=2\).
04

Solve for x

To find x for each equation, both sides of the equations can be taken the natural logarithm \(ln\). This will give \(x = ln(1)\) for the first equation and \(x = ln(2)\) for the second equation.
05

Calculate Decimal Approximations

For the final step, the decimal approximation for each solution can be found using calculator. Thus we obtain \(x \approx 0\) for the first solution and \(x \approx 0.69\) for the second solution, correct to two decimal places.

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Key Concepts

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

Natural Logarithms
Natural logarithms are an important concept in solving exponential equations. In algebra, when you have an equation involving the exponential function, such as \(e^x\), taking the natural logarithm usually helps isolate the variable \(x\). This is because natural logarithms (notated as \(\ln\)) are the inverse operation of exponentials with base \(e\).
  • If you have \(e^x = a\), you can solve for \(x\) by taking the natural logarithm of both sides to get \(x = \ln(a)\).
  • This step allows you to solve for \(x\) directly, which is especially helpful when dealing with complexities like those found in the transformation from quadratic equations involving exponential bases.
When solving problems such as the one provided, natural logarithms are used to deduce the value of \(x\) after back substituting the quadratic solutions into their exponential equivalents.
Quadratic Equation
The quadratic equation comes into play when transforming an exponential equation to a more familiar form. In the example given, the initial exponential equation \(e^{2x} - 3e^x + 2 = 0\) is transformed using substitution into a quadratic equation form. By letting \(e^x = y\), we restate the problem as \(y^2 - 3y + 2 = 0\). This makes it solvable using typical methods for quadratic equations.
  • A quadratic equation is of the form \(ay^2 + by + c = 0\), where \(a\), \(b\), and \(c\) are constants.
  • In this scenario, \(a=1\), \(b=-3\), and \(c=2\).
The solutions to the quadratic equation are found using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). By applying this formula, you find the roots that eventually help solve for \(x\) in the context of the original exponential equation.
Decimal Approximation
Decimal approximations are useful when exact solutions involve complex numbers or irrational numbers that cannot be easily expressed in a simpler form. They provide a practical way of presenting numerical solutions to a reasonable degree of accuracy, especially when using technology like calculators.In the exercise, after finding the logarithmic solutions \(x = \ln(1)\) and \(x = \ln(2)\), we aim for precision by converting these into decimal form:
  • \(x = \ln(1)\) translates to \(x \approx 0.00\) because \(\ln(1) = 0\).
  • \(x = \ln(2)\) results in \(x \approx 0.69\) when calculated to two decimal places.
This approach ensures that the final answers can be used in practical applications or further calculations where specific numerical values are required.
Substitution Method
The substitution method is a powerful tool that simplifies complex equations by temporarily replacing a difficult element with a simpler term. In the given exercise, substitution makes the exponential equation more manageable. The concept can be outlined as follows:
  • You encounter a challenging expression: \(e^{2x} - 3e^x + 2 = 0\).
  • We introduce a substitution \(e^x = y\), transforming it into \(y^2 - 3y + 2 = 0\), a straightforward quadratic equation.
  • This substitution reduces complexity, allowing easy application of algebraic methods like the quadratic formula to solve for \(y\).
  • After solving, replace the substitution back to its original form to find the solution in terms of the original variable.
Following this process aids in systematically breaking down and solving equations that initially appear daunting due to their format or expression.

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Most popular questions from this chapter

In Exercises \(1-40,\) use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \ln \sqrt[2]{x} $$

In Exercises \(1-40,\) use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log M^{-8} $$

The formula $$ t=\frac{1}{c}[\ln A-\ln (A-N)] $$ describes the time, \(t,\) in weeks, that it takes to achieve mastery of a portion of a task, where \(A\) is the maximum learning possible, \(N\) is the portion of the learning that is to be achieved, and \(c\) is a constant used to measure an individual's learning style. a. Express the formula so that the expression in brackets is written as a single logarithm. b. The formula is also used to determine how long it will take chimpanzees and apes to master a task. For example, a typical chimpanzce learning sign language can master a maximum of 65 signs. Use the form of the formula from part (a) to answer this question: How many weeks will it take a chimpanzee to master 30 signs if \(c\) for that chimp is \(0.03 ?\)

In Exercises \(41-70\), use properties of logarithms to condense each logarithmic expression. Write the expression as a single logarithm whose coefficient is \(I .\) Where possible, evaluate logarithmic expressions. $$ 4 \ln (x+6)-3 \ln x $$

In Exercises \(1-40,\) use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log _{5}(7 \cdot 3) $$
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