Problem 43
Question
Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ e^{2 x}-3 e^{x}+2=0 $$
Step-by-Step Solution
Verified Answer
The solution set in terms of natural logarithms is \(x = \ln(2)\) and \(x = 0\). The decimal approximation is \(x = 0.69\) and \(x = 0\).
1Step 1: Factoring the Expression
First, let \(u = e^{x}\). Therefore, the equation can be rewritten as \(u^{2}-3u+2=0\). Factoring this quadratic gives \((u-2)(u-1)=0\).
2Step 2: Solving for \(u\)
Setting each factor equal to zero gives us \(u = 2\) and \(u = 1\). Recall that \(u = e^{x}\), so \(e^{x} = 2\) and \(e^{x} = 1\).
3Step 3: Expressing in term of Natural Logarithms
Now, take the natural logarithm of both sides of these equations to express in terms of natural logarithms. Therefore, we have \(x = \ln(2)\) and \(x = \ln(1)\). The second equation simplifies to \(x = 0\) since natural logarithm of 1 equals zero.
4Step 4: Decimal Approximation
Using a calculator, \(\ln(2)\) to two decimal places is \(0.69\). Thus, the decimal approximations of \(x\) are \(0.69\) and \(0\).
Key Concepts
Natural LogarithmsQuadratic FactoringDecimal Approximation
Natural Logarithms
Natural logarithms are an essential concept in mathematics, particularly when dealing with exponential equations like the one mentioned in the exercise. The natural logarithm, denoted as \(\ln(x)\), is the inverse operation of raising the base of the natural logarithm, \(e\), to a power. In other words, if you have \(e^y = x\), then \(y = \ln(x)\).
Understanding this relationship is crucial because it allows us to solve for x when we have equations involving \(e\). In the context of the exercise, once the exponential equation is factored and reduced to simpler forms \(e^{x} = 2\) and \(e^{x} = 1\), the next natural step is to take the natural logarithm of both sides. This process effectively 'unlocks' x from the exponent, making it possible to solve for x directly.
Additionally, knowing that \(\ln(1) = 0\) simplifies calculations and is a key point that students should remember. Natural logarithms are used extensively in various fields of study such as science, engineering, and finance, making them an invaluable tool in the toolbox of mathematics.
Understanding this relationship is crucial because it allows us to solve for x when we have equations involving \(e\). In the context of the exercise, once the exponential equation is factored and reduced to simpler forms \(e^{x} = 2\) and \(e^{x} = 1\), the next natural step is to take the natural logarithm of both sides. This process effectively 'unlocks' x from the exponent, making it possible to solve for x directly.
Additionally, knowing that \(\ln(1) = 0\) simplifies calculations and is a key point that students should remember. Natural logarithms are used extensively in various fields of study such as science, engineering, and finance, making them an invaluable tool in the toolbox of mathematics.
Quadratic Factoring
Quadratic factoring is a technique used to break down quadratic expressions into simpler, multipliable factors. In general, a quadratic equation has the form \(ax^2 + bx + c = 0\), where a, b, and c are constants. The equation from the exercise, once we substitute \(u = e^{x}\), resembles this form with \(u^2 - 3u + 2 = 0\).
The goal of factoring is to rewrite the quadratic as a product of two binomials: \(u^2 - 3u + 2 = (u - 2)(u - 1)\). Finding these factors is akin to solving a puzzle where the product of the last terms must equal the constant term (c) and the sum of the inner and outer products must give us the middle term (-3u in this case).
Through factoring, we decompose the problem into simpler parts that are easier to solve: \(u - 2 = 0\) and \(u - 1 = 0\). This method is fundamental to solving quadratic equations and understanding it can significantly enhance a student's problem-solving skills. It's also a precursor to more advanced algebra concepts such as completing the square and using the quadratic formula.
The goal of factoring is to rewrite the quadratic as a product of two binomials: \(u^2 - 3u + 2 = (u - 2)(u - 1)\). Finding these factors is akin to solving a puzzle where the product of the last terms must equal the constant term (c) and the sum of the inner and outer products must give us the middle term (-3u in this case).
Through factoring, we decompose the problem into simpler parts that are easier to solve: \(u - 2 = 0\) and \(u - 1 = 0\). This method is fundamental to solving quadratic equations and understanding it can significantly enhance a student's problem-solving skills. It's also a precursor to more advanced algebra concepts such as completing the square and using the quadratic formula.
Decimal Approximation
Decimal approximation is the process of converting exact numbers into approximated values to a specific number of decimal places, often for practical use or ease of communication. In mathematics, particularly when solving equations with irrational solutions or in the presence of logarithms, it becomes necessary to use a calculator to find a decimal approximation.
In our exercise, after applying natural logarithms to solve for x (\(x = \ln(2)\) and \(x = \ln(1)\)), we need to translate these logarithmic answers into decimal form. This is where a calculator is handy; it gives us the approximated values, which are easier to grasp and use in real-world applications. For instance, \(\ln(2)\) is approximately \(0.69\) when rounded to two decimal places. Such approximations are especially beneficial in fields like science and engineering, where precise measurements are often conveyed in decimal form.
Furthermore, understanding how to round numbers correctly and to what degree of accuracy is needed in various contexts is a fundamental skill in mathematics. Remember to round up or down based on the number in the third decimal place and beyond to ensure accuracy up to the second decimal place, as shown in the exercise's solution.
In our exercise, after applying natural logarithms to solve for x (\(x = \ln(2)\) and \(x = \ln(1)\)), we need to translate these logarithmic answers into decimal form. This is where a calculator is handy; it gives us the approximated values, which are easier to grasp and use in real-world applications. For instance, \(\ln(2)\) is approximately \(0.69\) when rounded to two decimal places. Such approximations are especially beneficial in fields like science and engineering, where precise measurements are often conveyed in decimal form.
Furthermore, understanding how to round numbers correctly and to what degree of accuracy is needed in various contexts is a fundamental skill in mathematics. Remember to round up or down based on the number in the third decimal place and beyond to ensure accuracy up to the second decimal place, as shown in the exercise's solution.
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