Problem 7 - Contemporary Precalculus

Question

Solve the equation without using logarithms. $$9^{x^{2}}=3^{-5 x-2}$$

Step-by-Step Solution

Answer

Question: Solve the equation $$9^{x^{2}}=3^{-5 x-2}$$ without using logarithms. Answer: \(x=\frac{-1}{2}\) and \(x=-2\)

Step 1: Rewrite the bases as powers of 3

Since 9 is equal to \(3^2\), the left side of the equation can be rewritten as \((3^2)^{x^2}\), and the right side stays as it is. So, we obtain; $$(3^2)^{x^2}=3^{-5x-2}$$

Step 2: Use the power of a power property

Using the power of a power property, \((a^b)^c=a^{bc}\), we are able to simplify the expression: $$3^{2x^2}=3^{-5x-2}$$

Step 3: Compare the exponents

Since the bases are the same, we can set the exponents equal; $$2x^2=-5x-2$$

Step 4: Rearrange and solve for x

This is a quadratic equation. Let's rearrange it: $$2x^2+5x+2=0$$ Now we can try to factor this equation: $$(2x+1)(x+2)=0$$ This gives us two possible solutions: $$2x+1=0 \Longrightarrow x=\frac{-1}{2}$$ and $$x+2=0 \Longrightarrow x=-2$$ So, the solutions to the equation are \(x=\frac{-1}{2}\) and \(x=-2\).

Key Concepts

Exponent Properties

Understanding exponent properties is crucial when solving exponential equations. An exponent, also known as a power, represents the number of times a base number is multiplied by itself. For example, in the expression \(3^4\), 3 is the base and 4 is the exponent, indicating that 3 should be multiplied by itself 4 times.

When dealing with exponential equations, certain properties can simplify the process:

• The Power of a Power property states that \( (a^b)^c = a^{bc} \) where \( a \) is the base, and \( b \) and \( c \) are the exponents.
• The Product of Powers property indicates that when multiplying two exponents with the same base, you can add the exponents: \( a^b \cdot a^c = a^{b+c} \).
• The Quotient of Powers property allows for the subtraction of exponents when dividing: \( \frac{a^b}{a^c} = a^{b-c} \) when \( a \) is not zero.
• Zero Exponent means that any base \( a \) (except zero) raised to the power of zero is 1: \( a^0 = 1 \).
• The Negative Exponent rule indicates that \( a^{-b} = \frac{1}{a^{b}} \), for \( a \) not equal to zero.

Exploiting these properties allows for considerable simplification when solving exponential equations and will enable students to transform equations into a more manageable form, as shown in the example equation provided in the exercise.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, meaning the highest exponent of the variable (usually \( x \) ) is 2. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \) , where \( a \) , \( b \) , and \( c \) are constants, and \( a \eq 0 \).

These equations can be solved by various methods, including:

• Factoring, where the equation is expressed as a product of its factors.
• Completing the square, which involves creating a perfect square trinomial from the quadratic equation.
• The quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which is a universal method that can be used when the equation cannot be factored easily.
• Graphing, where the values of x where the graph crosses the x-axis represent the solutions to the equation.

To apply any of these methods effectively, a clear understanding of the properties of quadratic equations is paramount. For instance, the process of factoring relies on recognizing patterns and relationships between the coefficients and constants within the equation.

Factoring Quadratic Equations

Factoring is a method used to solve quadratic equations, which can be particularly effective when the equation is easily decomposable into a product of binomials. To factor a quadratic equation like \( ax^2 + bx + c = 0 \), one must find two numbers that both add up to \( b \) and multiply to \( ac \). When \( a \) is equal to 1, factoring becomes more straightforward as you only need to look for factors of \( c \) that add up to \( b \).

However, when \( a \) is not 1, the process is known as factoring by grouping, which requires an extra step of finding a pair of numbers that work for \( ac \) and then using them to split the middle term and factor by grouping. Once factored, the equation turns into \( (ax + m)(x + n) = 0 \), where \( m \) and \( n \) are the numbers found through factoring. Applying the Zero Product Property, which states that if a product equals zero, then at least one of the factors must be zero, we can set each binomial equal to zero to solve for x:

• \(\ ax + m = 0 \)
• \(\ x + n = 0 \)

Following these steps will reveal the solutions to the original quadratic equation. This method is especially useful when dealing with equations like the one presented in the exercise, where straightforward factors can be found.