Problem 52 - Precalculus Mathematics for Calculus

Question

Solve the logarithmic equation for \(x .\) $$\log x+\log (x-3)=1$$

Step-by-Step Solution

Answer

The solution is \( x = 5 \).

Step 1: Use the Product Rule of Logarithms

Recall that the sum of two logarithms with the same base can be expressed as a single logarithm by multiplying the arguments. So, \( \log(x) + \log(x - 3) \) becomes \( \log(x(x - 3)) \). Thus, the equation becomes \( \log(x(x - 3)) = 1 \).

Step 2: Convert Logarithmic Equation to Exponential Form

Since \( \log_{10}(A) = B \) is equivalent to \( 10^B = A \), we convert the logarithmic equation \( \log(x(x - 3)) = 1 \) to the exponential form, which gives us \( x(x - 3) = 10^1 \) or \( x(x - 3) = 10 \).

Step 3: Expand and Rearrange the Quadratic Equation

Expand the expression \( x(x - 3) \) to \( x^2 - 3x \). Then set the equation \( x^2 - 3x = 10 \) to \( x^2 - 3x - 10 = 0 \) by subtracting 10 from both sides.

Step 4: Factor the Quadratic Equation

To factor the quadratic \( x^2 - 3x - 10 = 0 \), we look for two numbers that multiply to -10 and add to -3. These numbers are -5 and 2. Thus, we factor the equation as \( (x - 5)(x + 2) = 0 \).

Step 5: Solve for x

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

Step 6: Check for Extraneous Solutions

Substitute each solution back into the original equation to check for validity. If \( x = 5 \), \( \log(5) + \log(2) = \log(10) = 1 \) holds true. If \( x = -2 \), \( \log(-2) \) is undefined as there are no real log values for negative numbers. Therefore, \( x = -2 \) is not a valid solution.

Key Concepts

Product Rule of Logarithms

Logs can sometimes be tricky, but the product rule of logarithms simplifies solving equations like this one by combining two logs into one. Whenever you see an expression like

• \(\log(a) + \log(b)\),

you can merge them into a single log expression:

• \(\log(a \cdot b)\).

Why does this work? Logs, at their core, help us tackle powers of numbers. This rule is rooted in the property that allows multiplication inside the log to correspond to addition of separate logs. In our exercise, \(\log(x) + \log(x - 3)\) cleverly combines into \(\log(x(x - 3))\). This simplification paves the way toward easily solving the equation.

Exponential Form Conversion

Switching between logarithmic and exponential forms is often essential when solving log equations. Remember, a log equation like

• \(\log_{10}(A) = B\),

translates to an exponential equation:

• \(10^B = A\).

This conversion bridges the gap between logarithms and regular numbers, allowing us to work with well-understood algebraic methods. In this specific example, \(\log(x(x - 3)) = 1\) becomes \(x(x - 3) = 10\). This step celebrates the beauty in how exponential notation offers a new, clearer perspective on our equation. Consider exponential notation the tool that helps dismantle logs into more manageable pieces.

Factoring Quadratic Equations

Quadratic equations often pop up in math exercises. The goal? Finding values of \(x\) that satisfy the equation. For \(x(x - 3) = 10\), we first expand to \(x^2 - 3x - 10 = 0\). Next, factoring breaks it down into parts that are easier to solve. To accomplish this, think: which numbers multiply to -10 and add to -3? The magic numbers are -5 and 2! Therefore, you factor the quadratic as

• \((x - 5)(x + 2) = 0\)

This factorization step transforms a single complex equation into simpler ones, setting the stage for finding possible solutions.

Extraneous Solutions

Solutions including negative logs can be sneaky. Although mathematically correct, they can be invalid when solving equations like this logarithmic one. That's because logarithms of negative numbers are undefined in the context of real numbers. When we solved

• \((x - 5)(x + 2) = 0\), yielding solutions \(x = 5\) and \(x = -2\).

The solution \(x = -2\) was an "extraneous solution". A solution in a mathematical sense, but not applicable to our problem. An extra check by substitution verifies valid solutions, crucial in ensuring complete and correct solutions to the equation. NEVER skip this verification step—it confirms our only valid answer here is \(x = 5\).