Problem 40 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the laws of logarithms to solve the equation. $$\log (x+7)-\log (x-2)=1$$

Step-by-Step Solution

Answer

The short answer to the given problem is: Apply the logarithm property to rewrite the equation as \(\log\left(\frac{x+7}{x-2}\right)=1\). Then convert to exponential form, obtaining \(10^1 = \frac{x+7}{x-2}\). Next, simplify the equation and obtain \(10x-20=x+7\). Finally, solve for x to get \(x=3\).

Step 1: Use the logarithm property

Rewrite the equation using the logarithm property \(\log a - \log b = \log (\frac{a}{b})\). By applying this property to the given equation, we'll have: \(\log (x+7)-\log (x-2)=\log\left(\frac{x+7}{x-2}\right)=1\)

Step 2: Convert to exponential form

Now, let's convert this logarithmic equation to its equivalent exponential form using the property \(\log_b a=c\) can be written as \(b^c=a\). In this case, since we have the base 10 logarithm, the exponential form of the equation will be: \(10^1 = \frac{x+7}{x-2}\)

Step 3: Simplify the equation

To solve for \(x\), let's simplify the exponential equation by multiplying both sides by the denominator of the fraction: \(10 (x-2) = x+7\) Now, distribute 10 to each term inside the parentheses: \(10x - 20 = x + 7\)

Step 4: Solve for x

To solve for \(x\), let's first move the term containing the variable to the left side of the equation by subtracting \(x\) from both sides: \(10x - x = 7 + 20\) Simplify: \(9x = 27\) Finally, divide both sides by 9 to isolate x: \(x = 3\) The solution to the given equation is \(x=3\).

Key Concepts

Laws of Logarithms

Logarithms have fascinating mathematical properties that make solving complex equations simpler. The laws of logarithms are essential tools. They help transform expressions using easier or more familiar forms. In this exercise, the difference form of the laws is particularly useful:

• Log Difference Law: This law states that \( \log a - \log b = \log \left( \frac{a}{b} \right) \). It allows you to combine two logarithms into one. This is what was initially applied to transform the given problem.

By understanding and using these laws effectively, you can turn a seemingly complicated problem into a manageable one.

Solving Equations

Solving logarithmic equations often requires strategic thinking to isolate the variable. After applying the logarithmic law, we had:\[ \log \left( \frac{x+7}{x-2} \right) = 1 \]To solve such equations, converting the logarithmic form to an exponential form is usually the next step. This shifts the equation from a logarithmic to a linear setup, where traditional solving methods can apply. Following this, you simplify the equation by handling all terms systematically. Always ensure variable terms end up on one side, allowing you to solve efficiently by adding, subtracting, multiplying, or dividing. The steps in this exercise show a classic method to tackle and simplify equations. Persistence and attention to detail are vital to ensure no step or term is overlooked.

Exponential Form

The transition from logarithmic equations to exponential form is pivotal. It involves rewriting the equation, which might initially seem intimidating, into something familiar.

• For base 10 logarithms: \( \log_{10} a = c \) converts to \( 10^c = a \).

In our example, turning the equation \( \log \left( \frac{x+7}{x-2} \right) = 1 \) into \( 10^1 = \frac{x+7}{x-2} \) immediately simplifies the path to finding solutions. This form is helpful because it removes the log and lets us work directly with the numbers. Therefore, recognizing when and how to convert between logarithmic and exponential forms is a critical skill.

Mathematical Properties

Understanding mathematical properties forms the foundation for solving equations like the one in this exercise. Here are some fundamental properties used in solving our problem:

• Distributive Property: Important when simplifying expressions, like distributing \( 10 \) over \( (x-2) \) to form a linear equation.
• Equality Property: Ensures operations applied to one side are also applied to the other, maintaining balance and consistency.

Apply these properties methodically, making sure each transformation gets you closer to isolating the variable. By doing so, equations become less of a challenge and more of an exercise in logical consistency. Familiarity with these properties equips you to solve even as equations become more complex.