Problem 48 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ \left(\log _{10} 2 x\right)^{2}=\log _{10}(2 x)^{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2}\) and \(x = 50\).

Step 1: Introduce Simplification

Notice that both sides of the equation involve logarithms. Start by simplifying the right-hand side of the equation \(\log_{10}(2x)^2\). According to the logarithmic power rule, \(\log_b(a^n) = n \cdot \log_b(a)\). This means \(\log_{10}(2x)^2 = 2 \cdot \log_{10}(2x)\).

Step 2: Rewrite the Equation

Rewriting the given equation using the simplification from Step 1 results in \( (\log_{10} 2x)^2 = 2 \cdot \log_{10} 2x \).

Step 3: Let Variable Substitute

Let's substitute \( y = \log_{10}(2x) \). This transforms the equation into \( y^2 = 2y \).

Step 4: Set Equation to Zero

Rearrange the equation to set it to zero: \( y^2 - 2y = 0 \).

Step 5: Factor the Equation

Factor the quadratic equation \( y^2 - 2y = 0 \), which gives \( y(y - 2) = 0 \).

Step 6: Solve for y

Set each factor equal to zero and solve for \( y \). Thus, \( y = 0 \) or \( y = 2 \).

Step 7: Back-Substitution

Substitute back \( y = \log_{10}(2x) \) to find \( x \). Thus, \((\log_{10}(2x) = 0)\) leads to \(2x = 1\) so \(x = \frac{1}{2}\). Similarly, \((\log_{10}(2x) = 2)\) leads to \(2x = 100\) so \(x = 50\).

Step 8: Conclusion

Therefore, the solutions to the equation are \(x = \frac{1}{2} \) or \(x = 50 \).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions, which means they help us determine what exponent we need to raise a base to, in order to obtain a certain number. In simple terms, if you have a number, a base, and you know that the base raised to some power equals the number, the logarithm helps us find this power. The expression \(log_b(a)\) is asking "What power should \(b\) be raised to get \(a\)?" This involves the base \(b\) and the argument \(a\). For example, \(log_{10}(100) = 2\), because \(10^2 = 100\). Logarithms have several important properties,

• Product Rule: \(log_b(M \, N) = \log_b(M) + \log_b(N)\)
• Quotient Rule: \(log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\)
• Power Rule: \(\log_b(M^n) = n \, \log_b(M)\). This rule is particularly useful in transforming and solving logarithmic equations as it simplifies exponential terms to multiplicative ones.

These properties allow us to manipulate and solve logarithmic equations just like we would solve algebraic equations, often leading to simplified expressions that are easier to handle.

Factoring

Factoring is a method used to solve quadratic equations where we write the equation as a product of its binomials. This method is particularly useful when the equation can be easily rewritten in a form where each term corresponds directly to the factors of a product that yields zero, because any number multiplied by zero results in zero itself.Let’s consider an example: if you have an equation \(y^2 - 2y = 0\), we can factor it as \(y(y - 2) = 0\). To solve this, set each factor equal to zero, solving each resulting simple equation separately:

• If \(y = 0\), then this satisfies the equation immediately.
• If \(y - 2 = 0\), solving gives \(y = 2\).

By applying factoring, not only do we simplify the process of finding solutions, but we also gain insight into the structure of the quadratic equation itself. Factoring is a quick and elegant way of handling quadratics, provided they can be easily decomposed into simpler expressions.

Quadratic Formula

The quadratic formula is a universal method for solving any quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. If factoring doesn’t simplify the equation, the quadratic formula is your go-to tool. The formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides a systematic way to find the roots or solutions of a quadratic equation. Here’s how each part works:

• \(-b\) reflects the equation about the \(y\)-axis.
• \(\pm \sqrt{b^2 - 4ac}\) accounts for the nature of the roots:
  • If the term under the square root, the discriminant \((b^2 - 4ac)\), is positive, the equation has two distinct real solutions.
  • If it is zero, there is exactly one real solution (called a double root).
  • If it is negative, the solutions are complex numbers, indicating no real intercepts.
• \(\frac{1}{2a}\) scales the solution according to the leading coefficient \(a\).

The quadratic formula works for every quadratic equation, providing a fail-safe method for precisely determining the solutions, ensuring no potential roots are overlooked.