Problem 72 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log \frac{2-5 x}{2(x+8)}=0 $$

Step-by-Step Solution

Answer

The exact solution is \( x = -2 \).

Step 1: Understand the Equation

The given equation is \( \log \frac{2-5x}{2(x+8)} = 0 \). This means that the logarithm of the expression \( \frac{2-5x}{2(x+8)} \) compared to base 10 is equal to zero. If a logarithm is zero, then the argument inside must be 1, since \( \log(1) = 0 \).

Step 2: Set the Argument Equal to 1

To solve the equation, set the argument equal to 1: \[ \frac{2-5x}{2(x+8)} = 1 \].This equation states that the fraction inside the logarithm is equal to 1.

Step 3: Solve the Fraction Equation

Multiply both sides of the equation \( \frac{2-5x}{2(x+8)} = 1 \) by \( 2(x+8) \) to eliminate the fraction:\( 2-5x = 2(x+8) \).

Step 4: Expand and Simplify

Expand the equation \( 2(x+8) \) on the right side:\( 2-5x = 2x + 16 \). Now, simplify this equation: \( 2 - 5x = 2x + 16 \).

Step 5: Isolate Variable x

To isolate \( x \), move terms involving \( x \) to one side and constants to the other:Add \( 5x \) to both sides: \( 2 = 7x + 16 \).Subtract 16 from both sides:\( 2 - 16 = 7x \), which simplifies to \( -14 = 7x \).

Step 6: Solve for x

Divide both sides of the equation \( -14 = 7x \) by 7 to solve for \( x \): \( x = -2 \).

Step 7: Verify the Solution

Substitute \( x = -2 \) back into the original argument of the logarithm to ensure it's valid:\( \frac{2-5(-2)}{2(-2+8)} = \frac{2 + 10}{2 \times 6} = \frac{12}{12} = 1 \).Since this equals 1, the solution is valid.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in algebra, where we determine the value of a variable that makes the equation true. In this exercise, we are dealing with a logarithmic equation: \( \log \frac{2-5x}{2(x+8)} = 0 \). A crucial property of logarithms is that when they equal zero, the expression inside must equal one, because \( \log(1) = 0 \).

To solve it, we first apply this property and set the argument equal to one: \( \frac{2-5x}{2(x+8)} = 1 \). The goal is to isolate \( x \) by unraveling the fraction.

• Multiply both sides by \( 2(x+8) \) to eliminate the fraction.
• Simplify the equation step by step, leaving \( x \) by itself.
These steps ensure the equation is solved correctly, guiding you to the accurate answer.

Algebra

Algebra plays a vital role in solving equations, particularly when dealing with expressions that must be simplified and manipulated. In the given problem, algebraic techniques unravel the expression inside the logarithm to make it easier to work with. After multiplying to eliminate the fraction, the equation becomes linear: \( 2-5x = 2(x+8) \).

The next step in algebra is expanding the terms using the distributive property, which gives us \( 2-5x = 2x + 16 \). This step is essential. By getting all \( x \) terms on one side of the equation and constants on the other, a clear path to solving for \( x \) is established. The resulting equation can be tackled straightforwardly using the rules of equivalent expressions to isolate \( x \).

• Add or subtract terms to both sides to consolidate like terms.
• Divide by the coefficient of \( x \) to solve for its value.

This systematic approach in algebra ensures logical progression towards finding the solution.

Mathematical Verification

Verifying the solution in mathematics is crucial to confirm the accuracy of your results. After solving the equation and finding \( x = -2 \), we must ensure this value is correct by substituting it back into the original expression. This step checks whether the solution is valid and meaningful within the context of the scenario.

Substitute \( x = -2 \) into the argument \( \frac{2-5(-2)}{2(-2+8)} \):
\[ \frac{2 + 10}{12} = \frac{12}{12} = 1 \]

• The result is 1, aligning perfectly with the initial requirement for the logarithm: \( \log(1) = 0 \).
• This substitution confirms that \( x = -2 \) is indeed a valid and correct solution.

This validation step is key in mathematics as it assures not only that the equation is correctly solved but also that logical consistency is maintained.