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Magazine Chapter 6: Problem 133
Solve: \(3^{2 x-1}-4 \cdot 3^{x}+9=0\)
Short Answer
Expert verified
\( x = 1 \) or \( x = 2 \)
Step by step solution
01
Express the Equation in Terms of a Single Variable
Substitute a new variable for the common base expression. Let鈥檚 set \( y = 3^x \). Then the equation becomes \( 3^{2x-1} - 4 \times 3^x + 9 = 0 \). Notice that \( 3^{2x-1} = \frac{(3^x)^2}{3} = \frac{y^2}{3} \).
02
Rewrite the Equation
Using the substitution from Step 1, rewrite the original equation in terms of \( y \):\( \frac{y^2}{3} - 4y + 9 = 0 \).
03
Multiply by 3 to Clear the Fraction
To eliminate the fraction, multiply each term by 3:\( y^2 - 12y + 27 = 0 \).
04
Solve the Quadratic Equation
Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for solving \( y^2 - 12y + 27 = 0 \). Here, \( a=1, b=-12, \text{ and } c=27 \). Calculate the discriminant, \( b^2 - 4ac = 144 - 108 = 36 \).
05
Find the Roots of the Quadratic Equation
Substitute back into the quadratic formula to find:\( y = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} \).Thus, \( y = \frac{18}{2} = 9 \) or \( y = \frac{6}{2} = 3 \). So, \( y = 9 \) or \( y = 3 \).
06
Back-Substitute to Find \( x \)
Recall that \( y = 3^x \). So, if \( y = 9 \), then \( 3^x = 9 \) which gives \( x = 2 \). If \( y = 3 \), then \( 3^x = 3 \) which gives \( x = 1 \).
07
Verify the Solutions
Substitute \( x=1 \) and \( x=2 \) back into the original equation to verify both are correct solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
exponential functions
Exponential functions are mathematical expressions of the form \(a^x\), where \(a\) is a positive constant called the base, and \(x\) is the exponent. They exhibit rapid growth or decay depending on whether the base is greater than 1 or between 0 and 1, respectively. In the given exercise, the exponential function is \(3^{2x-1}\), which involves the base 3 raised to the power of an expression involving the variable \(x\). Understanding exponential functions is crucial as they appear in many areas of mathematics and real-world applications, such as population growth, radioactive decay, and financial modeling.
substitution method
The substitution method is a powerful algebraic technique used to simplify complex equations by substituting one part of the equation with a new variable. This method helps reduce an equation to a more manageable form. In our exercise, we let \( y = 3^x \), transforming the original exponential equation into a simpler polynomial form. This substitution makes it easier to apply algebraic methods to solve the equation, such as converting it into a quadratic equation that can be solved with standard techniques.
quadratic equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving quadratic equations is a fundamental skill in algebra. In our problem, after substituting \(y\to 3^x\) and rearranging terms, we obtain a quadratic equation: \(y^2 - 12y + 27 = 0\). There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. The quadratic formula, \(y = \frac{-b \,\pm \, \sqrt{b^2 - 4ac}}{2a}\), is often the most straightforward approach as it works for all quadratic equations.
discriminant
The discriminant is a part of the quadratic formula and is found inside the square root: \(b^2 - 4ac\). It provides important information about the nature of the roots of a quadratic equation without needing to find them explicitly. The discriminant can tell us the following:
In our exercise, the discriminant \(b^2 - 4ac = 36\) is positive, indicating two distinct real roots for the quadratic equation \(y^2 - 12y + 27 = 0\). Substituting this back into the quadratic formula then leads us to the solutions.
- If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real roots.
- If \(b^2 - 4ac = 0\), there is exactly one real root (also called a repeated or double root).
- If \(b^2 - 4ac < 0\), the quadratic equation has two complex roots.
In our exercise, the discriminant \(b^2 - 4ac = 36\) is positive, indicating two distinct real roots for the quadratic equation \(y^2 - 12y + 27 = 0\). Substituting this back into the quadratic formula then leads us to the solutions.
