Problem 53 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log (2 x-5)-\log (x-3)=1$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{25}{8} \).

Step 1: Apply the properties of logarithms

Start by using the property of logarithms that states \( \log a - \log b = \log \left( \frac{a}{b} \right) \). This allows us to combine the two logarithms in the equation. The equation becomes: \[ \log \left( \frac{2x-5}{x-3} \right) = 1 \]

Step 2: Exponentiate to eliminate the logarithm

To eliminate the logarithm and solve for \( x \), exponentiate both sides of the equation using base 10. This is because the logarithm given is base 10 (common logarithm). Thus, the equation becomes: \[ \frac{2x-5}{x-3} = 10^1 \] which simplifies to: \[ \frac{2x-5}{x-3} = 10 \]

Step 3: Solve for x

Clear the fraction by multiplying both sides by \( x-3 \): \[ 2x - 5 = 10(x - 3) \] Simplify and expand the right side: \[ 2x - 5 = 10x - 30 \] Rearrange the equation to isolate terms involving \( x \) on one side: \[ 2x - 10x = -30 + 5 \] \[ -8x = -25 \] Divide both sides by -8 to solve for \( x \): \[ x = \frac{25}{8} \]

Step 4: Verify the solution

Substitute \( x = \frac{25}{8} \) back into the original equation to verify: \[ \log\left(2\left(\frac{25}{8}\right) - 5\right) - \log\left(\frac{25}{8} - 3\right) = 1 \] Simplifying: \[ \log\left(\frac{50}{8} - \frac{40}{8}\right) - \log\left(\frac{25}{8} - \frac{24}{8}\right) \] \[ \log\left(\frac{10}{8}\right) - \log\left(\frac{1}{8}\right) \] \[ \log\left(10\right) = 1 \] which holds true, confirming the solution.

Key Concepts

Logarithm Properties

Logarithms are incredibly useful in mathematics, especially when dealing with exponential relationships. They have several properties that simplify equations and calculations. One such property is the difference of logarithms, expressed as \( \log a - \log b = \log \left( \frac{a}{b} \right) \). This is the quotient rule for logarithms. It allows us to combine two separate logarithmic terms into one, by transforming a subtraction into a division inside the log function.

• Logarithms convert multiplication into addition:
• \( \log(a \times b) = \log a + \log b \).
• Convert division into subtraction:
• \( \log \left( \frac{a}{b} \right) = \log a - \log b \).
• Raise powers inside a logarithm:
• \( \log(a^b) = b \cdot \log a \).

Using these properties simplifies complex calculations and is a cornerstone in algebraic manipulations, particularly when dealing with logarithmic equations.

Exponentiation

Exponentiation is the process of raising a number to a power. In the context of solving logarithmic equations, exponentiation is often used to eliminate the logarithmic function from the equation. Given the base of the logarithm (often 10 for common logs), you can exponentiate both sides to "cancel out" the logarithm:

• If \( \log_b(a) = c \), then by exponentiating, \( a = b^c \).

In our problem, we exponentiated both sides with base 10 to remove the logarithm, freeing the equation to be solved algebraically. Effective use of exponentiation allows us to transition from the logarithmic form back to a more familiar algebraic one. This process is crucial when you need to find exact values in logarithmic equations.

Solving Equations

Solving equations requires isolating the variable of interest and finding its value. There are several techniques involved, from simplifying expressions to using operations like addition, subtraction, multiplication, or division.

• Identify forms: Examine the equation to find logarithms, exponents, or linear terms.
• Use algebraic manipulations: Apply operations or properties (like inverse functions) systematically to isolate the variable.
• Simplify: Combine like terms and reduce fractions where possible.
• Verify: Substitute your solution back into the original equation to check accuracy.

In our example, after simplifying using logarithmic and exponentiation properties, we needed to clear fractions by multiplying. Rearranging terms continued until we isolated \( x \). Always finish with substitution to confirm your solution is correct and satisfies the original equation.