Problem 92 - Algebra for College Students

Question

Solve each equation. $$ \log _{10} x+\log _{10}(x-3)=1 $$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Apply Logarithm Property

Use the logarithmic identity \(\log_b (m) + \log_b (n) = \log_b (mn)\). This means we can rewrite the given equation as \(\log_{10} (x(x-3)) = 1\).

Step 2: Exponential Form

Convert the logarithmic equation to exponential form using the definition \(\log_b (a) = c\) means \(b^c = a\). Hence, \(10^1 = x(x-3)\).

Step 3: Simplify the Equation

Simplify the equation to \(x^2 - 3x = 10\).

Step 4: Rearrange to a Quadratic Equation

Rearrange the equation to standard quadratic form: \(x^2 - 3x - 10 = 0\).

Step 5: Solve the Quadratic Equation

Factor the quadratic equation. We have \(x^2 - 3x - 10 = (x - 5)(x + 2) = 0\).

Step 6: Find the Roots

Set each factor equal to zero: \(x - 5 = 0\) and \(x + 2 = 0\). This gives us \(x = 5\) and \(x = -2\).

Step 7: Validate the Solutions

Check the potential solutions in the original logarithmic expressions. Note that \(\log_{10}\) is only defined for positive arguments. \(x = -2\) is not valid as it results in negative arguments for \(\log\). So, \(x = 5\) is the only valid solution.

Key Concepts

Logarithmic Properties

Logarithms are powerful tools in mathematics, especially when it comes to simplifying complex equations. One of the most vital properties is the product rule of logarithms. It states that if you have two logs with the same base and they're added together, you can combine them. This looks like \[\log_b (m) + \log_b (n) = \log_b (mn) \]This property greatly simplifies equations with multiple log terms, making them easier to solve.

• In the original problem, we had \(\log_{10} x + \log_{10} (x-3)\).
• Using the property, these terms combine to form \(\log_{10} (x(x-3))\).

Understanding logarithmic properties allows for the reduction of expressions, turning complex equations into simpler ones that are more manageable.

Exponential Form

Converting logarithmic equations into exponential form is a key step in finding solutions. Remember the basic definition of a logarithm: \[\log_b (a) = c \] can be rewritten as \[b^c = a \] This transformation is useful for solving equations involving logs.Once the expression \(\log_{10} (x(x-3)) = 1\) was simplified, we converted it into exponential form:

• Since the base here is 10 and the result of the logarithm is 1, we convert it to \(10^1 = x(x-3)\).

This step transitions the problem into a polynomial equation, which can be further simplified and solved.

Quadratic Equation

After converting into exponential form, the equation becomes a quadratic polynomial. Here's how:

• The equation \(10 = x(x-3)\) simplifies to \(x^2 - 3x = 10\).
• Subtract 10 from both sides to get the standard quadratic form: \(x^2 - 3x - 10 = 0\).

Recognizing and solving quadratic equations is critical. Quadratics are usually solved by factoring, completing the square, or using the quadratic formula. In our case, factoring provided a clean solution:

• \(x^2 - 3x - 10 = (x - 5)(x + 2) = 0\).
• This gives potential roots \(x = 5\) and \(x = -2\).

Grasping quadratic forms and their solutions plays a crucial role in solving algebraic expressions.

Solution Validation

Validating solutions is an essential step to ensure they are correct and applicable in context, especially when logarithms are involved. The constraint that \[\log_{10}\] must have positive arguments brings clarity in validation.

• Check the solutions \(x = 5\) and \(x = -2\).
• For \(x = -2\), the argument of the logarithms become negative which isn't allowed, so this solution is not valid.
• \(x = 5\) works perfectly since both \(\log_{10} 5\) and \(\log_{10} (5-3)\) are defined and positive.

Thus, the only valid solution to our logarithmic equation is \(x = 5\). Always verify your potential solutions in the context of the original problem to avoid errors.