Problem 83 - Intermediate Algebra

Question

For Problems \(81-97\), solve each of the equations. $$ \log _{10} x+\log _{10}(x-21)=2 $$

Step-by-Step Solution

Answer

The solution is \(x = 25\).

Step 1: Use the Product Rule of Logarithms

Recall that the sum of logarithms with the same base can be combined using the product rule: \(\log_b a + \log_b c = \log_b (a \times c)\). Apply this rule here to combine the logarithms: \(\log_{10} x + \log_{10} (x-21) = \log_{10} [x(x-21)]\).

Step 2: Simplify the Equation

Using the product rule from the previous step, rewrite the equation as \(\log_{10}(x(x-21)) = 2\).

Step 3: Transform to Exponential Form

To eliminate the logarithm, rewrite the logarithmic equation in its exponential form. Since \(\log_{10} a = b\) implies \(a = 10^b\), you can write \(x(x-21) = 10^2\).

Step 4: Solve the Quadratic Equation

Simplify the exponential equation: \(x^2 - 21x = 100\). Rearrange it to form a standard quadratic equation: \(x^2 - 21x - 100 = 0\).

Step 5: Factor the Quadratic Equation

Try to factor the quadratic equation \(x^2 - 21x - 100 = 0\) into two binomials. Look for two numbers whose product is -100 and sum is -21. These numbers are -25 and 4, so factor as \((x - 25)(x + 4) = 0\).

Step 6: Solve for x

Use the zero-product property on the factored form: \((x - 25) = 0\) or \((x + 4) = 0\). Solving these gives \(x = 25\) or \(x = -4\).

Step 7: Verify the Solutions

Check each solution in the original logarithmic equation. Substitute \(x = 25\) and \(x = -4\) back into \(\log_{10} x + \log_{10}(x-21) = 2\). Since \(\log_{10}(-4)\) is undefined, the only valid solution is \(x = 25\).

Key Concepts

Quadratic Equations

Quadratic equations are remarkable because they are polynomial equations of degree two. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). In this form, \(a\), \(b\), and \(c\) are constants, with \(aeq0\). Here, solving a quadratic equation typically involves combining strategies such as factoring, completing the square, or using the quadratic formula.
One of the encounters of quadratic equations in solving logarithmic equations is from transforming logarithmic expressions. For the given problem, after using the product rule of logarithms, we transform the problem into a quadratic form by eliminating the logarithm and then expanding: \(x(x - 21) = 100\). Thus, rearranging it results in \(x^2 - 21x - 100 = 0\).
Solving quadratic equations often leads us to find zero-product property solutions which provide the roots of the equation. Each solution gives the values that satisfy the original equation when substituted back.

Product Rule of Logarithms

The product rule of logarithms is an essential property used to simplify the sum of logarithmic expressions. Remember, when you encounter \(\log_b a + \log_b c\), you can apply the product rule, which combines them into a single logarithm: \(\log_b (a \times c)\).
In our problem, we started with two separate logarithms, \(\log_{10} x\) and \(\log_{10}(x-21)\). Using the product rule allowed us to transform this sum into a single logarithmic expression \(\log_{10} [x(x-21)]\), making the equation simpler to work with.
This transformation is crucial as it compacts our equation and leads us to handle it much easier through the subsequent steps. By applying this rule, we reduce the complexity, aiding in an effective solution process.

Exponential Form

To solve logarithmic equations, converting them to exponential form is a powerful tool. A logarithmic equation, like \(\log_b a = c\), can be transformed into its exponential form \(a = b^c\).
In the context of our problem, we utilize this property to eliminate the logarithm. Once the product rule of logarithms gives us \(\log_{10}(x(x-21)) = 2\), the next step is to convert this into an exponential form. Applying the conversion yields \(x(x-21) = 10^2\).
This transformation simplifies our task of solving the equation drastically, moving from a logarithmic form to a polynomial equation, which is often easier to handle and solve.

Zero-Product Property

The zero-product property is a fundamental rule used in algebra to solve equations. It states that if the product of two factors is zero, then at least one of the factors must be zero. This property is effectively used to find solutions for quadratic equations.
Once a quadratic equation is factored into binomials, like \((x-25)(x+4) = 0\), the zero-product property allows us to set each factor equal to zero separately. This method leads to individual solutions: \(x - 25 = 0\) gives \(x = 25\), and \(x + 4 = 0\) gives \(x = -4\).
After applying the zero-product property, always ensure to verify each solution, especially when dealing with logarithmic equations. This is because certain values might render parts of the original equation undefined (like \(\log_{10}(-4)\) in this problem). Hence, only \(x = 25\) is valid for our problem.