Problem 78 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log (3-2 x)-\log (x+9)=0 $$

Step-by-Step Solution

Answer

The exact solution is \( x = -2 \). No decimal approximation needed as the solution is an integer.

Step 1: Apply the logarithmic subtraction rule

Recognize that the equation given is a subtraction of logarithms: \( \log (3-2x) - \log (x+9) = 0 \). Apply the subtraction rule \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Thus, rewrite the equation as \( \log \left( \frac{3-2x}{x+9} \right) = 0 \).

Step 2: Remove logarithm by applying the exponentiation property

Since \( \log(\frac{3-2x}{x+9}) = 0 \), according to the exponentiation property of logarithms, \( \frac{3-2x}{x+9} = 10^0 = 1 \).

Step 3: Solve the rational equation

The equation \( \frac{3-2x}{x+9} = 1 \) implies that the numerator must equal the denominator. Set \( 3 - 2x = x + 9 \).

Step 4: Simplify and solve for x

Rearrange the equation: \( 3 - 2x = x + 9 \) to \( 3 - 9 = x + 2x \). This simplifies to \( -6 = 3x \). Divide both sides by 3 to find \( x = -2 \).

Step 5: Verify the solution

We need to check if \( x = -2 \) satisfies the original equation or constraints. Substitute \( x = -2 \) back into the expressions inside the logarithms: \( 3 - 2(-2) = 7 \) and \( -2 + 9 = 7 \). Since both are positive, the log values are defined, and the solution is correct.

Key Concepts

Logarithmic properties

Logarithmic properties are key rules that allow us to manipulate and simplify expressions involving logarithms. Understanding these properties is essential when solving logarithmic equations. In our exercise, we primarily use the subtraction rule which states:

• \( \log a - \log b = \log \left( \frac{a}{b} \right) \)

This property helps us rewrite a difference of two logarithms as a single logarithmic statement of a division between the inner quantities of the logs. By using this rule, we can convert and simplify the given equation \( \log (3-2x) - \log (x+9) = 0 \) into \( \log \left( \frac{3-2x}{x+9} \right) = 0 \).
Another important property used is the exponentiation property, which allows us to eliminate the logarithm by transforming the equation \( \log(y) = p \) into \( y = 10^p \), assuming that the logarithm is base 10. In our problem, this transforms \( \log \left( \frac{3-2x}{x+9} \right) = 0 \) into \( \frac{3-2x}{x+9} = 1 \).
These properties are crucial for breaking down complex logarithmic expressions into more manageable parts, facilitating easier manipulation and solution of the equation.

Solving rational equations

Rational equations involve expressions that are fractions where the numerator and/or the denominator contain a variable. Solving these kinds of equations requires eliminating the fraction by algebraically manipulating the equation.
Once we simplify the logarithmic equation to \( \frac{3-2x}{x+9} = 1 \), our task is to solve this rational equation. We start by recognizing that since the fractional expression must equal 1, the numerator must be equal to the denominator. Therefore, we set:

• \( 3 - 2x = x + 9 \)

Next, we rearrange the terms:

• \( 3 - 9 = x + 2x \)

Simplifying gives us \( -6 = 3x \), and by dividing both sides by 3, we determine:

• \( x = -2 \)

This solution shows that once the equation is simplified, solving a rational equation often involves straightforward algebra to isolate the variable.

Checking solutions in equations

It is crucial to check solutions in logarithmic and rational equations to confirm their validity, especially since certain values can cause undefined expressions. When solving our equation, we found that\( x = -2 \). We need to ensure this solution does not cause any negative arguments inside the logarithms or any other mathematical contradiction.
To verify, we substitute \( x = -2 \) back into the original expressions:

• For \( 3 - 2(-2) \), we find \( 7 \)
• For \( -2 + 9 \), we find \( 7 \)

Both results yield positive values, meaning the original logarithmic expressions are defined and well-behaved. This confirms that our solution \( x = -2 \) is correct.
Always remember to check any constraints related to the domain of the equation, especially in logarithmic functions, to ensure all parts of your solution are valid.