Question

Solve each equation. $$ 9^{2 x} \cdot 27^{x^{2}}=3^{-1} $$

Short answer

The solutions are \( x = -\frac{1}{3} \) and \( x = -1 \).

Step-by-step solution

Express each base as a power of 3

Rewrite 9 and 27 as powers of 3. The number 9 can be written as \(3^2\), and the number 27 can be written as \(3^3\). So the equation becomes: \[ (3^2)^{2x} \times (3^3)^{x^2} = 3^{-1} \]

Simplify using exponent rules

Apply the power of a power rule: \[ (3^2)^{2x} = 3^{4x} \] and \[ (3^3)^{x^2} = 3^{3x^2} \]. Update the equation to: \[ 3^{4x} \times 3^{3x^2} = 3^{-1} \]

Combine the exponents

Since the bases are the same, add the exponents on the left-hand side: \[ 3^{4x + 3x^2} = 3^{-1} \]

Set the exponents equal

Because the bases are the same, the exponents must be equal: \[ 4x + 3x^2 = -1 \]

Solve the quadratic equation

Rearrange the equation to standard quadratic form: \[ 3x^2 + 4x + 1 = 0 \]. This can be solved using the quadratic formula: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \]. In our case, a = 3, b = 4, and c = 1.Substituting into the formula: \[ x = \frac{{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot 1}}}{2 \cdot 3} = \frac{{-4 \pm \sqrt{16 - 12}}}{6} = \frac{{-4 \pm 2}}{6} \]

Simplify solutions

Solve for the two possible values of x: \[ x = \frac{{-4 + 2}}{6} = \frac{{-2}}{6} = -\frac{1}{3} \] and \[ x = \frac{{-4 - 2}}{6} = \frac{{-6}}{6} = -1 \]

Key concepts

quadratic equations

Quadratic equations are a type of polynomial equation of the second degree. They have the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not equal to zero. Solving a quadratic equation typically involves finding the values of \( x \) that make the equation true. There are several methods to solve quadratic equations, such as factoring, completing the square, and using the quadratic formula.

In solving quadratic equations, the quadratic formula is a powerful tool that applies to all forms of quadratic equations. The formula is given by: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \]. By inserting the coefficients of your specific quadratic equation (in this case, 3, 4, and 1 from \( 3x^2 + 4x + 1 = 0 \)), you can determine the solutions for \( x \).

Additionally, understanding how to rearrange the terms of a quadratic equation into its standard form is crucial. For instance, converting \( 4x + 3x^2 = -1 \) to \( 3x^2 + 4x + 1 = 0 \) helps in applying the quadratic formula effectively.

exponent rules

Exponent rules are essential when working with exponential equations as they help simplify complex expressions. Some vital exponent rules include:

• Power of a Power Rule: \( (a^m)^n = a^{mn} \)
• Product of Powers Rule: \( a^m \cdot a^n = a^{m+n} \)
• Quotient of Powers Rule: \( \frac{a^m}{a^n} = a^{m-n} \)
• Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \)

These rules help in expressing complex expressions in their simplest forms. In our example, we used the power of a power rule to express \( (3^2)^{2x} \) as \( 3^{4x} \) and \( (3^3)^{x^2} \) as \( 3^{3x^2} \). We also used the product of powers rule to combine the exponents: \( 3^{4x} \cdot 3^{3x^2} = 3^{4x + 3x^2} \). Finally, applying the rule of like bases, we equated the exponents to solve for \( x \).

base conversion

Base conversion is the process of expressing numbers in terms of different bases. It allows us to simplify and solve equations involving powers and exponents more efficiently. Common base conversions include turning numbers like 9 and 27 into powers of a more manageable base, such as 3. For instance:

• The number 9 can be written as \( 3^2 \)
• The number 27 can be written as \( 3^3 \)

In our problem, we converted 9 and 27 to their base 3 equivalents to simplify the equation. This gave us \( (3^2)^{2x} \cdot (3^3)^{x^2} = 3^{-1} \), which could then be simplified using exponent rules. Base conversion is crucial for simplifying the bases to allow the use of exponent rules effectively.By using a common base, it becomes straightforward to combine and compare the exponents, leading to simpler equations that are easier to solve.