Problem 78 - 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 approximation needed.

Step 1: Apply Logarithm Properties

The given equation is \( \log(3-2x) - \log(x+9) = 0 \). Use the logarithm property \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \) to combine the logarithms: \[ \log\left(\frac{3-2x}{x+9}\right) = 0 \]

Step 2: Eliminate the Logarithm

Since \( \log(y) = 0 \) implies that \( y = 1 \), set the inside of the logarithm to equal 1: \[ \frac{3-2x}{x+9} = 1 \]

Step 3: Solve the Resulting Equation

Cross-multiply to solve for \( x \): \[ 3-2x = x+9 \] Rearrange the equation to solve for \( x \): \[ 3 - 9 = x + 2x \] \[ -6 = 3x \] \[ x = -2 \]

Step 4: Verify the Solution

To verify, substitute \( x = -2 \) back into the original expressions inside the logarithms: \[ \log(3 - 2(-2)) = \log(3 + 4) = \log(7) \] \[ \log((-2) + 9) = \log(7) \] Both expressions are equal, thus \( x = -2 \) is a valid solution.

Key Concepts

Properties of Logarithms

Logarithms have special properties that make it easier to manipulate and solve equations involving them. One especially useful property is the difference of logarithms:

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

This allows a subtraction of logarithms to be rewritten as a single logarithm, with a fraction inside. For our problem, this property was used to combine \( \log(3-2x) - \log(x+9) = 0 \) into \( \log\left(\frac{3-2x}{x+9}\right) = 0 \) making it much simpler to solve.Using logarithmic properties can help us simplify the problem and apply further algebraic methods to find the solution. The key takeaway is the power of these properties to transform equations into more manageable forms, which can then be tackled with basic algebra.

Exact Solutions

Finding exact solutions in logarithmic equations allows us to understand the precise values that satisfy the equation. Once we simplify an equation using logarithm properties, we often end up with an equation without logarithms. After setting the expression within the logarithm to 1 (since \( \log(y) = 0 \) implies \( y = 1 \)), we obtained:

• \( \frac{3-2x}{x+9} = 1 \)

This can be rid of the fraction by cross-multiplying, leading to a simple linear equation:

• \( 3 - 2x = x + 9 \)

Solving it, step-by-step, involves:

• Rearranging terms to \( 3 = 3x + 9 \)
• Subtracting to simplify \( -6 = 3x \)
• Dividing both sides by 3 to find \( x = -2 \)

Arriving at \( x = -2 \) gives the exact solution to our logarithmic equation.

Approximations of Solutions

While some solutions to logarithmic equations can be expressed exactly, others may require approximations—often useful for practical applications. In this particular problem, since the exact solution \( x = -2 \) was easily derived, an approximation wasn't necessary. However, approximations usually come into play when dealing with more complex solutions or when a numerical answer is more helpful. Calculating approximations to several decimal places can provide a clearer numerical insight:

• Use a calculator to determine values
• Round to the needed precision
• For instance, solutions might require approximation up to four decimal places

Approximations become essential when an exact form is difficult, or when measuring the solution's impact on a real-world scenario.

Verification of Solutions

Verifying a solution involves checking that it satisfies the original equation. This step ensures no errors were made in the calculation and that the solution fits all conditions of the problem.In the given problem, after finding \( x = -2 \), we substitute it back into each logarithmic part of the original equation:

• \( \log(3 - 2 \times (-2)) = \log(3 + 4) = \log(7) \)
• \( \log((-2) + 9) = \log(7) \)

Since both evaluations result in \( \log(7) \), the solution is verified as correct.Verification helps confirm the exactness of a solution and ensures it adheres to the constraints provided by the logarithms. Always perform this step to ascertain the validity of your solution, especially after intricate manipulations or simplifications.