Problem 18 - Intermediate Algebra

Question

For Problems \(15-22\), solve each logarithmic equation. $$ \log (2 x-1)-\log (x-3)=1 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{29}{8} \).

Step 1: Use Logarithm Properties

Apply the property of logarithms that states \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \). Here you have \( \log (2x - 1) - \log (x - 3) = \log \left( \frac{2x-1}{x-3} \right) \).

Step 2: Exponential Form

Rewrite the equation in exponential form. The equation \( \log \left( \frac{2x-1}{x-3} \right) = 1 \) indicates that \( \frac{2x-1}{x-3} = 10^1 \). Therefore, \( \frac{2x-1}{x-3} = 10 \).

Step 3: Solve for x

Cross multiply to solve for \( x \):\[2x - 1 = 10(x - 3)\]Simplify and solve the equation:\[2x - 1 = 10x - 30\]Re-arrange the terms:\[2x - 10x = -30 + 1\]\[-8x = -29\]Divide both sides by -8:\[x = \frac{29}{8}\]

Step 4: Verify the Solution

Substitute \( x = \frac{29}{8} \) back into the original equation to verify:\(\log \left(2\left(\frac{29}{8}\right)-1\right) - \log \left(\frac{29}{8} - 3\right) = 1\)Calculate inside the logarithms:\(\log \left(\frac{50}{8} - \frac{8}{8}\right) - \log \left(\frac{29}{8} - \frac{24}{8}\right) = 1\)\(\log \left(\frac{42}{8}\right) - \log \left(\frac{5}{8}\right)\)Which simplifies to:\(\log \left(42/5\right)\)Since exponential and logarithmic forms match, \( x = \frac{29}{8} \) is correct.

Key Concepts

Logarithm Properties

Logarithms are powerful mathematical tools that transform multiplicative relationships into additive ones. One essential property used in solving logarithmic equations is the difference of logs. This property states:

• \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \)

In simpler terms, when you subtract one logarithm from another, you can rewrite it as a single logarithm of a fraction. This concept is helpful in collapsing multiple logs into one, making it easier to manipulate equations.
In the original exercise, this property is applied to simplify \( \log (2x - 1) - \log (x - 3) = \log \left( \frac{2x-1}{x-3} \right) \). This transformation prepares the equation for further simplification.

Exponential Form

Converting a logarithmic equation into its exponential form is a key step in solving these types of problems. The idea here is to express the equation in terms of powers of the base of the logarithm. This step leverages the fundamental relationship between logarithms and exponents:

• If \( \log_a b = c \), then \( b = a^c \)

In the given problem, we have \( \log \left( \frac{2x-1}{x-3} \right) = 1 \). This can be rewritten in exponential form as:

• \( \frac{2x-1}{x-3} = 10^1 \)
• Which simplifies to \( \frac{2x-1}{x-3} = 10 \)

This conversion is crucial because it transforms the logarithmic equation into a more straightforward algebraic form that can be readily solved through typical means.

Cross Multiplication

Once a logarithmic equation is expressed in exponential form, the next step typically involves solving an algebraic fraction. One effective method for handling fractions of this nature is cross multiplication.
Cross multiplication allows you to eliminate the denominators by multiplying across the equal sign, essentially cross-multiplying the terms:

• For \( \frac{A}{B} = C \), cross multiply to obtain \( A = B \, C \)

For the exercise, cross multiplication turns \( \frac{2x-1}{x-3} = 10 \) into:

• \( 2x - 1 = 10(x - 3) \)

This equation can then be simplified through basic algebraic operations, namely expansion, combination of like terms, and division to isolate and find \( x \). Cross multiplication thus serves as a bridge from a complex fraction to a manageable linear equation.

Solution Verification

After solving the equation to find the value of \( x \), it's crucial to verify that the solution is correct. Solution verification ensures that the obtained solution satisfies the original equation and that no mistakes were made during the manipulation.
This is done by substituting the found \( x \) back into the original logarithmic expression:

• Initial verification involves ensuring that each term inside the logarithms results in a positive value, as logarithms of negative numbers are undefined.
• If the original equation includes \( \log (b) - \log (c) \), substitute \( b \) and \( c \) with their respective expressions using the solved \( x \).

For the example in focus, substitute \( x = \frac{29}{8} \) back, compute the values inside the logs, and simplify. Successfully ending up with the original constant (1, in this case) reaffirms that the solution is correct. This step not only confirms the process but also reinforces the correctness of applied concepts and arithmetic.