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Magazine Chapter 13: Problem 16
Solve the given equations. $$8^{x}=4^{x^{2}-1}$$
Short Answer
Expert verified
The solutions are \(x = 2\) and \(x = -\frac{1}{2}\).
Step by step solution
01
Rewrite Bases with Common Base
To solve the equation \(8^x = 4^{x^2-1}\), we first rewrite both sides of the equation with the common base of 2. Note that \(8 = 2^3\) and \(4 = 2^2\). So our equation becomes \((2^3)^x = (2^2)^{x^2-1}\).
02
Apply the Power Rule
Using the power rule \((a^m)^n = a^{m\cdot n}\), transform the equation: \((2^3)^x = 2^{3x}\) and \((2^2)^{x^2-1} = 2^{2(x^2-1)} = 2^{2x^2-2}\). Our equation is now \(2^{3x} = 2^{2x^2-2}\).
03
Equate the Exponents
Since the bases are the same, we can equate the exponents: \(3x = 2x^2 - 2\). This gives us the equation \(2x^2 - 3x - 2 = 0\).
04
Use the Quadratic Formula
The equation \(2x^2 - 3x - 2 = 0\) is a quadratic equation. The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can be used, where \(a = 2\), \(b = -3\), and \(c = -2\). Calculate the discriminant: \(b^2 - 4ac = (-3)^2 - 4(2)(-2) = 9 + 16 = 25\).
05
Solve for "x"
Using the discriminant and quadratic formula: \(x = \frac{-(-3) \pm \sqrt{25}}{2(2)} = \frac{3 \pm 5}{4}\). This gives us two solutions for \(x\): \(x = \frac{8}{4} = 2\) and \(x = \frac{-2}{4} = -\frac{1}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Common Base Method
The common base method is a technique used to simplify exponential equations. It involves rewriting different bases to a single common base.
This technique is especially useful when dealing with exponential equations where each side of the equation has a different base.
This technique is especially useful when dealing with exponential equations where each side of the equation has a different base.
- Identify the common base: Find a common base to which each term can be converted. For example, in the equation \(8^x = 4^{x^2-1}\), the numbers 8 and 4 can both be expressed with base 2, since \(8 = 2^3\) and \(4 = 2^2\).
- Rewrite the equation: Transform all terms to the common base. This simplifies comparison and calculation. You'll rewrite the original equation as \((2^3)^x = (2^2)^{x^2-1}\).
Applying the Power Rule in Exponential Equations
The power rule is an essential algebraic property that enables us to simplify and manipulate complex exponential expressions. It states that \((a^m)^n = a^{m \cdot n}\), which allows us to rewrite and solve equations involving exponents.
To apply the power rule successfully, follow these steps:
To apply the power rule successfully, follow these steps:
- Simplify each side: In our example, using the power rule, \((2^3)^x\) becomes \(2^{3x}\) and \((2^2)^{x^2-1}\) becomes \(2^{2(x^2-1)} = 2^{2x^2-2}\).
- Equate the exponents: Once the bases are identical, you can set the exponents equal: \(3x = 2x^2 - 2\).
Using the Quadratic Formula
The quadratic formula is a key tool for solving quadratic equations, such as \(ax^2 + bx + c = 0\). It provides a systematic way to find solutions for these types of equations by using the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our step where we equated the exponents to get \(2x^2 - 3x - 2 = 0\), the quadratic formula becomes useful. Here are the steps to solve it:
- Identify coefficients: Find \(a = 2\), \(b = -3\), and \(c = -2\).
- Calculate the discriminant: The discriminant \(b^2 - 4ac\) determines the nature of the roots. Here, it complements the quadratic formula by showing real and distinct solutions, since \((-3)^2 - 4 \cdot 2 \cdot (-2) = 25\).
- Solve for x: Substitute into the quadratic formula to find solutions. For this example, \(x = \frac{3 \pm 5}{4}\) yields \(x = 2\) and \(x = -\frac{1}{2}\).
