Problem 51 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log x+\log (x-21)=2$$

Step-by-Step Solution

Answer

The solution is \( x = 25 \).

Step 1: Use the Product Rule for Logarithms

The equation is \( \log x + \log (x-21) = 2 \). According to the product rule, \( \log a + \log b = \log(ab) \). So, we can combine the logs to get \( \log(x(x-21)) = 2 \).

Step 2: Convert the Logarithmic Equation to Exponential Form

Using the property of logarithms that \( \log_b a = c \) means \( a = b^c \), we convert the equation to exponential form. Here, \( \log_{10}(x(x-21)) = 2 \) becomes \( x(x-21) = 10^2 \).

Step 3: Simplify the Exponential Equation

Solve \( x(x-21) = 100 \). Expanding gives \( x^2 - 21x = 100 \). Rearranging gives the quadratic equation \( x^2 - 21x - 100 = 0 \).

Step 4: Solve the Quadratic Equation

We solve the quadratic equation \( x^2 - 21x - 100 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -21 \), and \( c = -100 \).

Step 5: Calculate the Discriminant and Solutions

Calculate the discriminant: \( b^2 - 4ac = (-21)^2 - 4 \cdot 1 \cdot (-100) = 441 + 400 = 841 \). Since 841 is a perfect square, calculate \( x \):\[x = \frac{21 \pm \sqrt{841}}{2}\]\[= \frac{21 \pm 29}{2}\]This results in two solutions: \( x = \frac{50}{2} = 25 \) and \( x = \frac{-8}{2} = -4 \).

Step 6: Evaluate the Validity of Solutions

Since a logarithm cannot take negative or zero arguments, we check the solutions. The solution \( x = 25 \) works because both \( 25 \) and \( 25 - 21 = 4 \) are positive numbers. The solution \( x = -4 \) is invalid because you cannot take \( \log(-4) \) or \( \log(-25) \).

Step 7: Verify Using a Calculator

Substitute \( x = 25 \) back into the original equation. Calculate \( \log(25) + \log(4) \). The sum should equal \( 2 \) since \( \log_{10}(100) = 2 \). Using a calculator, verify that \( \log(25) + \log(4) \approx 2 \). The values confirm the solution.

Key Concepts

Product Rule for Logarithms

Understanding the product rule for logarithms is crucial when working with logarithmic equations. Logarithms have specific properties that make them easier to manipulate and solve when combined. The product rule states that the logarithm of a product is equal to the sum of the logarithms of its factors. In simpler terms:

• \( \log_a(xy) = \log_a x + \log_a y \)

When you have an equation like \( \log x + \log (x - 21) = 2 \), you can apply this rule to combine the logarithms on the left-hand side into a single logarithmic expression:

• \( \log(x(x-21)) \)

Applying the product rule simplifies the equation significantly, making the process of finding solutions more straightforward. Once the equation \( \log(x(x-21)) = 2 \) is established, solving for \( x \) becomes more manageable with exponential forms.

Exponential Form Conversion

Converting from logarithmic to exponential form is another key concept necessary for solving logarithmic equations. It helps bridge the gap between a logarithmic expression and a more solvable algebraic form. The basic idea is:

• If \( \log_b a = c \), then \( a = b^c \)

For the given equation \( \log(x(x-21)) = 2 \), we recognize that the base of the logarithm is 10 since it’s common logarithm (log base 10). Thus, we convert the equation to its exponential form:

• \( x(x-21) = 10^2 \)

This conversion transforms the logarithmic equation into a polynomial equation \( x^2 - 21x - 100 = 0 \), which is ready to be solved using algebraic methods like factoring or the quadratic formula. By utilizing exponential form conversion, you set the stage for efficiently solving the equation.

Quadratic Equation Solving

Once the logarithmic equation is converted to its exponential form, solving it often turns into solving a quadratic equation. In our case, the equation is \( x^2 - 21x - 100 = 0 \). Solving quadratic equations can be done through various methods:

• Factoring: Look for two numbers that multiply to -100 and add to -21.
• Quadratic Formula: Use \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which applies to any quadratic equation \( ax^2 + bx + c = 0 \).

In our scenario, the discriminant \( b^2 - 4ac \) is computed as 841, which happens to be a perfect square. This indicates that the quadratic formula can provide exact solutions:

• \( x = \frac{21 \pm 29}{2} \)
• Simplifying gives solutions \( x = 25 \) and \( x = -4 \).

However, only \( x = 25 \) is valid since logarithmic expressions require positive arguments. Solving the quadratic equation furnishes the solution we seek, underscoring the importance of understanding quadratic methods.