Problem 89 - A Graphical Approach to College Algebra

Question

$$\text { The given equations are quadratic in form. Solve each and give exact solutions.}$$ $$\left(\log _{2} x\right)^{2}+\log _{2} x=2$$

Step-by-Step Solution

Answer

The exact solutions for \(x\) are \(2\) and \(\frac{1}{4}\).

Step 1: Recognize the Equation Form

The equation \((\log_{2}x)^{2} + \log_{2}x = 2\) is quadratic in terms of \(y = \log_{2}x\). Rewrite the equation as \(y^{2} + y - 2 = 0\), which is a standard quadratic equation in terms of \(y\).

Step 2: Apply the Quadratic Formula

The standard form of a quadratic equation is \(ay^2 + by + c = 0\). Here, \(a = 1\), \(b = 1\), and \(c = -2\). The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values to get \(y = \frac{-1 \pm \sqrt{1 + 8}}{2}\).

Step 3: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac = 1 + 8 = 9\). Since the discriminant is a perfect square, the solutions for \(y\) are real and rational.

Step 4: Solve for y

Substitute the discriminant into the quadratic formula: \(y = \frac{-1 \pm 3}{2}\). This gives the solutions \(y = 1\) and \(y = -2\).

Step 5: Back Substitute to Solve for x

Since \(y = \log_{2}x\), we have two equations: \(\log_{2}x = 1\) and \(\log_{2}x = -2\). Solve each equation separately.

Step 6: Solve \(\log_{2}x = 1\)

Convert the logarithmic equation \(\log_{2}x = 1\) to its exponential form: \(x = 2^{1} = 2\).

Step 7: Solve \(\log_{2}x = -2\)

Convert the logarithmic equation \(\log_{2}x = -2\) to its exponential form: \(x = 2^{-2} = \frac{1}{4}\).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. In simpler terms, if you know about exponential functions like \(2^x\), then the logarithm with base 2 would be \(\log_2(x)\). This means that the logarithm gives you the power that you need to raise 2 to get \(x\).
In the equation \((\log_{2}x)^{2} + \log_{2}x = 2\), we treat \(\log_{2}x\) as a single variable, often called \(y\) in such transformations. This simplifies complex expressions, making it easier to handle problems involving logarithms.

• To solve the equation \(\log_{2}x = y\), remember it means \(2^y = x\).
• Logarithmic equations can be rewritten as exponential equations by using their definitions.

Understanding logarithms is crucial in various math contexts, including algebra and calculus. When faced with logarithmic terms in an equation, converting them into exponential form can give a fresh perspective and often simplifies finding solutions.

Discriminant

The discriminant is a component of the quadratic formula, found within the square root, typically represented as \( b^2 - 4ac \). This value tells us crucial information about the nature of the roots of a quadratic equation.
For the equation \( y^2 + y - 2 = 0 \), the discriminant is \( 1^2 - 4 \cdot 1 \cdot (-2) = 9 \). Some important facts about the discriminant include:

• If the discriminant is positive, as it is in this case, the equation has two different real roots.
• If it equals zero, there is exactly one real root, also known as a repeated or double root.
• If it’s negative, the roots are complex and not real numbers.

In our solution, because the discriminant is a perfect square (9), we could easily find rational solutions. Recognizing the discriminant helps you predict the type of solutions without solving the entire equation, acting as a decision tool in mathematical problem-solving.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ay^2 + by + c = 0\). It's given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
To find the solutions for \(y^2 + y - 2 = 0\), identify the coefficients: \(a = 1\), \(b = 1\), \(c = -2\). Substitute these into the formula:
\[ y = \frac{-1 \pm \sqrt{9}}{2} \]
This simplifies to \(y = 1\) and \(y = -2\). The "\(\pm\)" symbol indicates that there are two solutions, one adding and one subtracting the square root.

• The quadratic formula works for any quadratic equation, regardless of the complexity.
• Always compute the discriminant first to assess whether the roots are real or complex.

Using the quadratic formula ensures you can systematically and efficiently find solutions for any standard quadratic equation. It's a foundational tool often revisited throughout advanced mathematics.