Problem 130 - Algebra and Trigonometry

Question

Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the equation's solution set. Verify this value by direct substitution into the equation. $$ \log (x-15)+\log x=2 $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log (x-15)+\log x=2\) is \(x = 20\)

Step 1: Simplify the Equation

By rule of logarithms, \(\log_b a + \log_b c = \log_b (a \cdot c)\), here base ‘b’ is 10. So, \(\log (x-15)+\log x=2\) becomes \(\log ((x-15)x) = 2\) or equivalently \(\log (x^2 -15x) = 2\)

Step 2: Remove Logarithm

To remove the logarithm on the left side of the equation, use the inverse property of logarithms \(b^{\log_b a} = a\). Given equation is now \(x^2 -15x = 10^2\), which further simplifies to \(x^2 -15x - 100 = 0\)

Step 3: Solve for x

Solve the quadratic equation \(x^2 -15x - 100 = 0\) by factoring: \((x-20)(x+5) = 0\). The solutions to this equation are \(x=20\) and \(x=-5\)

Step 4: Verify the Solution

Verify the solutions by substituting them back into the original equation. Keep in mind that the domain of the logarithmic function is \(x>0\), so \(x=-5\) would be discarded. For \(x=20\), the original equation becomes: \(\log (20-15) + \log 20 = 2\). This simplifies to \(\log 5 + \log 20 = 2\), and by the rules of logarithms becomes \(\log (5*20) = \log (100) = 2\). So the solution is valid.

Key Concepts

Logarithms

Logarithms are a fundamental concept in mathematics used to solve equations involving exponential relationships. A logarithm answers the question: "To what power must the base (usually 10 or e) be raised to produce a given number?" For example, in the expression \( \log_{10} 100 = 2 \), 10 must be raised to the power of 2 to equal 100.

In an equation like \( \log (x-15) + \log x = 2 \), we apply the logarithmic property: \( \log_b a + \log_b c = \log_b (a \cdot c) \). This allows us to combine two logarithmic terms into a single one as \( \log((x-15)x) \), simplifying the problem. The base, implicitly 10 in "log" notation, dictates the kind of logarithmic operation involved.

Understanding these basic properties of logarithms helps in manipulating and solving logarithmic equations effectively.

Quadratic Equations

Quadratic equations form a pivotal part of algebra and equations such as \( x^2 - 15x - 100 = 0 \) are encountered often. These equations have the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

To solve a quadratic equation, you can use several techniques:

• Factoring: Finding two numbers whose product equals \( c \) and sum equals \( b \).
• Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), useful when factoring is complex.
• Completing the square: Involves creating a perfect square trinomial on one side of the equation.

In our solution, we factored the quadratic \( (x-20)(x+5) = 0 \), yielding solutions of \( x=20 \) and \( x=-5 \). However, because of the logarithmic domain restrictions, only \( x=20 \) is valid.

Solution Sets

The solution set of an equation is an important mathematical concept representing all values that satisfy the equation. For \( \log(x-15) + \log x = 2 \), the solution set comprises the values of \( x \) that make the equation true.

After simplifying and solving the quadratic equation to get \( x=20 \) and \( x=-5 \), we must verify these solutions. Verification checks if substituting back into the original equation yields a true statement.

However, not all solutions from the factorization are viable due to the domain of logarithms. It's crucial here to consider only positive values, since the logarithm of a non-positive number is undefined. This reduces the solution set effectively and is a step often overlooked but crucial for accuracy.

Graphing Utilities

In modern mathematics solving processes, graphing utilities like graphing calculators or software (e.g., Desmos or GeoGebra) are invaluable. They provide visual insight into equations, especially when finding intersections between graphs.

Graphing can visually demonstrate solutions appearing where two curves intersect. For \( \log(x-15) + \log x = 2 \), graph each side of the equation to see their intersection points. The \( x \)-coordinate at these points corresponds to the potential solutions of the equation.

These tools can check work done algebraically or reveal mistakes in manual calculations. Besides offering quick results, they help hone our understanding of graphical relationships within equations, encouraging a deeper comprehension of mathematical concepts.