Question

Solve each equation. $$ 4^{x} \cdot 2^{x^{2}}=16^{2} $$

Short answer

The solutions are \(x = 2\) and \(x = -4\).

Step-by-step solution

- Rewrite the bases

Rewrite all terms in the equation with base 2. Remember that 4 can be written as \(2^2\) and 16 can be written as \(2^4\). Thus, the equation \(4^x \times 2^{x^2} = 16^2\) becomes \((2^2)^x \times 2^{x^2} = (2^4)^2\).

- Apply the power rule

Use the power rule \((a^m)^n = a^{mn}\). Applying this to the equation, we get \((2^{2x}) \times 2^{x^2} = 2^{8}\)

- Simplify the equation

Since the bases are the same, we can combine the exponents on the left side of the equation. This gives us \(2^{2x + x^2} = 2^8\)

- Equate the exponents

Because the exponents must be equal when the bases are equal, we can set the exponents of base 2 equal to each other. This results in the equation \(2x + x^2 = 8\).

- Solve the quadratic equation

Rearrange the quadratic equation to standard form: \(x^2 + 2x - 8 = 0\). Now solve for \(x\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = -8\).

- Apply the quadratic formula

Substituting the values into the quadratic formula, we get \(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}\).

- Find the solutions

Simplify the solutions: \(x = \frac{-2 + 6}{2} = 2\) and \(x = \frac{-2 - 6}{2} = -4\). Therefore, the solutions to the equation are \(x = 2\) and \(x = -4\).

Key concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is generally written in the form \(ax^2 + bx + c = 0\). To solve for \(x\), we use:

• \( x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } \)

First, identify the coefficients \(a, b, \) and \(c\) from the equation. For example, if the equation is \( x^2 + 2x - 8 = 0\), then \( a = 1, b = 2, \) and \( c = -8\). Plug these values into the quadratic formula and simplify step by step:

• Calculate the discriminant \( b^2 - 4ac \).
• Take the square root of the discriminant.
• Apply the \pm \ to get two values.
• Divide by \( 2a \).

Finally, simplify the two possible solutions to find the values of \( x \).

Exponent Rules

Understanding exponent rules is key when solving equations with exponential terms. The common rules include:

• Product of Powers Rule: \( a^m \cdot a^n = a^{m+n} \)
• Power of a Power Rule: \( (a^m)^n = a^{mn} \)
• Power of a Product Rule: \( (ab)^m = a^m \cdot b^m \)
• Base Conversion: Rewrite bases to the same number, like converting \(4 \) to \( 2^2 \).

For example, the equation \( 4^x \cdot 2^{x^2} = 16^2 \) becomes easier to solve when converting all terms to base \( 2 \) and applying the power rules. This leads to combining the exponents and solving a simpler equation.

Combining Like Terms

Combining like terms is essential for simplifying equations. Like terms are terms that have the same variable raised to the same power.In the example \( (2^{2x}) \cdot 2^{x^2} = 2^8 \), the terms on the left are combined as follows:

• Both terms have the base \(2\).
• Combine the exponents by adding them: \( 2x + x^2 \).

This gives us a new equation: \( 2^{2x + x^2} = 2^8 \). By keeping bases consistent and combining exponents, solving the equation becomes simpler and straightforward.

Base Conversion

Base conversion involves rewriting numbers in a different base to simplify calculations. This is particularly useful in solving exponential equations.For instance, in the given equation \(4^x \cdot 2^{x^2} = 16^2 \), convert \( 4 \) and \( 16 \) to base \( 2 \) because it matches the base of other terms:

• \4 = 2^2\
• \16 = 2^4\

After conversion, the equation becomes \ (2^2)^x \cdot 2^{x^2} = (2^4)^2\. Applying the power of a power rule simplifies this to a common base format \ 2^{2x} \cdot 2^{x^2} = 2^8\, making the next steps easier.