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Magazine Chapter 4: Problem 32
Solve the equation. $$e^{x}-12 e^{-x}-1=0$$
Short Answer
Expert verified
The solution is \(x = \ln(4)\).
Step by step solution
01
Recognize a substitution
Notice that the equation involves both \(e^{x}\) and \(e^{-x}\), which suggests a substitution to simplify the expression. Let \(y = e^{x}\). Consequently, \(e^{-x} = \frac{1}{y}\). Substitute these into the equation.
02
Substitute into the equation
Using the substitution \(y = e^{x}\) and \(e^{-x} = \frac{1}{y}\), the equation becomes:\[y - 12\left(\frac{1}{y}\right) - 1 = 0\] Simplify this into:\[y - \frac{12}{y} - 1 = 0\] Multiply the entire equation by \(y\) to eliminate the fraction:
03
Eliminate the fraction
Multiplying the equation \(y - \frac{12}{y} - 1 = 0\) by \(y\) gives:\[y^2 - 12 - y = 0\]Which simplifies further to:\[y^2 - y - 12 = 0\]
04
Solve the quadratic equation
Solve the quadratic equation \(y^2 - y - 12 = 0\) using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -1\), and \(c = -12\):\[y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}\]\[y = \frac{1 \pm \sqrt{1 + 48}}{2}\]\[y = \frac{1 \pm 7}{2}\]
05
Find the roots of the quadratic
The solutions are:\[y_1 = \frac{1 + 7}{2} = 4\]\[y_2 = \frac{1 - 7}{2} = -3\]Since \(y = e^x\) cannot be negative, we discard \(y = -3\). Thus, \(y = 4\) is the only valid solution.
06
Solve for x
Recall the substitution \(y = e^x\). Now set \(e^x = 4\). Solve for \(x\) with the natural logarithm:\[x = \ln(4)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique to simplify complex equations by introducing a new variable. It's often used when dealing with equations that involve complex expressions containing multiple instances of the same function.
For example, with exponential equations involving terms like \(e^{x}\) and \(e^{-x}\), substitution helps simplify the problem. In our example, by setting \(y = e^{x}\), we can also express \(e^{-x}\) as \(\frac{1}{y}\).
For example, with exponential equations involving terms like \(e^{x}\) and \(e^{-x}\), substitution helps simplify the problem. In our example, by setting \(y = e^{x}\), we can also express \(e^{-x}\) as \(\frac{1}{y}\).
- This substitution turns the original complex equation into a more manageable form. By reformulating \(e^{x} - 12 e^{-x} - 1 = 0\) as \(y - \frac{12}{y} - 1 = 0\), it removes exponentials and makes the equation easier to handle.
- The goal of substitution is always to simplify the equation into a form that is easier to solve, often into a polynomial or another known form.
- Remember, once you solve the simpler equation, use back-substitution to express the solution in terms of the original variable.
Quadratic Equations
A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
Let's break down this type of equation using our example, which after substitution became \(y^2 - y - 12 = 0\).
Let's break down this type of equation using our example, which after substitution became \(y^2 - y - 12 = 0\).
- The coefficients in this equation are \(a = 1\), \(b = -1\), and \(c = -12\).
- The solutions to a quadratic equation can be found using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- For the discriminant \(b^2 - 4ac\), if it is positive, there are two distinct real roots. If zero, one real root exists; if negative, no real roots are found.
- In this case, the discriminant \(b^2 - 4ac = 1 + 48 = 49\) was positive, ensuring two real roots: \(y_1 = 4\) and \(y_2 = -3\).
Natural Logarithm
The natural logarithm, denoted as \(\ln(x)\), is the inverse operation of exponentiation with base \(e\). It's a critical concept when solving exponential equations, as it allows us to reverse the power operation.
In our example, after determining \(y = 4\) for \(y = e^x\), we arrive at \(e^x = 4\). To find \(x\), we apply the natural logarithm:
In our example, after determining \(y = 4\) for \(y = e^x\), we arrive at \(e^x = 4\). To find \(x\), we apply the natural logarithm:
- Applying the natural logarithm: \(x = \ln(4)\), effectively solves \(e^x = 4\) because \(\ln(e) = 1\).
- Remember that the natural logarithm is only defined for positive numbers because the exponential function \(e^x\) is never zero or negative.
- Natural logarithms have practical roles in various fields, especially in continuous compounding in finance, and examining growth and decay processes.
