Problem 22 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation for \(x\). $$2^{2 x}-4 \cdot 2^{x}+4=0$$

Step-by-Step Solution

Answer

The solution to the given equation \(2^{2x} - 4 \cdot 2^x + 4 = 0\) is \(x = 1\).

Step 1: Identify the common base

Notice that every term in the equation has a base of 2. The common base we will work with is \(2^x\).

Step 2: Apply substitution method

Let \(y=2^x\). Substituting this in the given equation will create a quadratic equation. The equation will now look like: $$y^2 - 4y + 4 = 0$$

Step 3: Solve the quadratic equation

In order to solve the quadratic equation, we can factor it: $$(y-2)(y-2) = 0$$ This gives the solution for \(y\): $$y=2$$

Step 4: Replace the substitution variable with the original variable

Now we need to replace \(y\) with \(2^x\): $$2^x = 2$$

Step 5: Solve for x using logarithms

To solve for the variable \(x\), we can take the logarithm on both sides of the equation. We'll use the base 2 logarithm (log2) here: $$\log_2{(2^x)} = \log_2{(2)}$$ Using the property of logarithms, we have: $$x \cdot \log_2{(2)} = \log_2{(2)}$$ Since \(\log_2{(2)} = 1\), the equation simplifies to: $$x = 1$$ Therefore, the solution to the given equation is \(x = 1\).

Key Concepts

Substitution Method

The substitution method is a powerful and straightforward technique used to simplify complex equations. This method works by replacing a part of the equation with a new variable. This often results in a simpler equation that is easier to handle.
For instance, in an expression like \(2^{2x} - 4 \cdot 2^x + 4 = 0\), we can set \(y = 2^x\). This change transforms the equation into \(y^2 - 4y + 4 = 0\), which is a quadratic equation.

• Choose a variable to substitute that simplifies the complexity.
• Write the original equation in terms of the new variable.
• Solve the simpler equation.
• Replace the substitution variable back with the original terms to find the final solution.

By substituting, we reduce the equation to a simpler form. This makes solving equations, especially complex ones, much more feasible and less error-prone.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They take the form of \(ax^2+bx+c=0\). These equations are fundamental in algebra and arise frequently in various mathematical problems.
The quadratic equation we obtained from our substitution is \(y^2 - 4y + 4 = 0\). This can be factored as \((y - 2)(y - 2) = 0\), giving us the solution \(y = 2\).

• Common methods for solving quadratic equations include factoring, using the quadratic formula, or completing the square.
• Factoring works when the quadratic expression can be written as a product of binomials.
• The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is the most general method and applies to any quadratic equation.

Understanding how to manipulate and solve quadratic equations is crucial, as they often pop up in various contexts like physics, engineering, and economics.

Logarithms

Logarithms allow us to solve equations involving exponentials by transforming them into linear equations. This is especially helpful when dealing with exponential equations where direct calculation is not straightforward.
In our given problem, after obtaining \(2^x = 2\), we can employ logarithms to solve for \(x\). By taking the logarithm (base 2) of both sides of the equation, we utilize the property \(\log_2{(2^x)} = x \cdot \log_2{(2)}\). Knowing that \(\log_2{(2)} = 1\), we find that \(x = 1\).

• Logarithms are the inverses of exponential functions.
• They have various properties, such as \(\log_b{(b^x)}=x\) and \(\log_b{(mn)}=\log_b{m} + \log_b{n}\).
• By converting an exponential equation into a logarithmic format, we simplify the process of finding the exponent.

Mastering logarithms is essential for understanding the solutions to exponential equations and is widely applicable across mathematics and science disciplines.