Problem 46 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ 2^{2 x}-12\left(2^{x}\right)+35=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\ln 5}{\ln 2} \) and \( x = \frac{\ln 7}{\ln 2} \).

Step 1: Substitute variable

Let us introduce a substitution to simplify the equation. Set \( y = 2^x \). This gives us the equation \( y^2 - 12y + 35 = 0 \).

Step 2: Identify quadratic form

Notice that the equation \( y^2 - 12y + 35 = 0 \) is a quadratic equation in the standard form \( ay^2 + by + c = 0 \), where \( a = 1 \), \( b = -12 \), and \( c = 35 \).

Step 3: Factor the quadratic equation

To factor \( y^2 - 12y + 35 \), we search for two numbers that multiply to 35 and add to -12. These numbers are -5 and -7. Thus, we can rewrite the equation as \((y - 5)(y - 7) = 0\).

Step 4: Solve for y

Using the zero-product property, set each factor equal to zero: \( y - 5 = 0 \) or \( y - 7 = 0 \). Solving these, we get \( y = 5 \) or \( y = 7 \).

Step 5: Substitute back for x

Recall that \( y = 2^x \). Thus, \( 2^x = 5 \) or \( 2^x = 7 \).

Step 6: Solve for x

Solve for \( x \) by taking the logarithm of both sides:- For \( 2^x = 5 \), \( x = \log_2 5 = \frac{\ln 5}{\ln 2} \).- For \( 2^x = 7 \), \( x = \log_2 7 = \frac{\ln 7}{\ln 2} \).

Key Concepts

Logarithms

Logarithms are a powerful mathematical tool that help us solve equations involving exponents, like those found in the solution of the quadratic equation. When you have an equation in the form of \( 2^x = a \), you can use logarithms to find \( x \). Logarithms answer the question: "To what exponent must we raise a certain base to obtain a given number?"

Here's how you use them in practice:

• Change of Base Formula: To solve \( 2^x = 5 \), we use the logarithm with base 2: \( x = \log_2 5 \).
• Conversion to Natural Logarithms (ln): When calculating using a calculator that lacks specific base logs, we can convert: \( \log_2 5 = \frac{\ln 5}{\ln 2} \).
• In the context of this problem, we use \( x = \log_2 7 \) or \( x = \log_2 5 \), which effectively solve \( 2^x = 7 \) or \( 2^x = 5 \) respectively.

Logarithms allow us to untangle and simplify complex problems, transforming exponentiation issues into more manageable operations.

Factoring

Factoring is a crucial technique when dealing with quadratic equations. A quadratic equation typically takes the form \( ay^2 + by + c = 0 \), and the goal in factoring is to express this equation as a product of simpler expressions.

• Identify Constants: For the equation \( y^2 - 12y + 35 = 0 \), here \( a = 1 \), \( b = -12 \), and \( c = 35 \).
• Find Two Numbers: These numbers should multiply to give the constant term, 35, and sum to give the linear term, -12.
• Rewrite Equation: After identifying -5 and -7 as these two numbers, we express \((y - 5)(y - 7) = 0\).
• Solution by Zeros Property: Each factor equals zero providing solutions \( y = 5 \) and \( y = 7 \).

Factoring simplifies the process of solving quadratic equations, making it easier to find the roots.

Substitution Method

The substitution method is a strategic approach that simplifies complex equations, especially helpful in solving quadratic equations embedded within exponential structures.

Let's take a closer look:

• Replace Variables to Simplify: We start by recognizing the equation's complex structure, such as \( 2^{2x} - 12(2^x) + 35 = 0 \). By substituting \( y = 2^x \), the equation becomes simpler.
• This transforms the equation to a familiar quadratic form: \( y^2 - 12y + 35 = 0 \).
• The benefit of this method is reducing the complexity of the equation, making it more manageable for further techniques like factoring.
• Reverse the Substitution: Once solutions for \( y \) are found, substitute back to solve for the original variable \( x \). Here, \( 2^x = 5 \) or \( 2^x = 7 \).

The substitution method is an effective strategy for transforming challenging problems into solvable formats, providing clear pathways to the solution.