Problem 23 - Algebra for College Students

Question

Solve each logarithmic equation and express irrational solutions in lowest radical form. $$ \log (3 x-1)=1+\log (5 x-2) $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{21}{47} \).

Step 1: Apply Logarithm Properties

Use the property of logarithms that states: \( \log a - \log b = \log \left( \frac{a}{b} \right) \). We aim to isolate terms involving logs. So, subtract \( \log(5x-2) \) from both sides of the equation:\[ \log(3x-1) - \log(5x-2) = 1 \] which can be rewritten using the property as:\[ \log\left( \frac{3x-1}{5x-2} \right) = 1 \].

Step 2: Exponentiate Both Sides

Use the logarithmic identity that states if \( \log_a (b) = c \), then \( a^c = b \). With base 10, this transforms the equation to:\[ \frac{3x-1}{5x-2} = 10^1 = 10 \].

Step 3: Solve the Algebraic Equation

Now solve for \( x \) by eliminating the fraction. Multiply both sides by \( (5x-2) \):\[ 3x - 1 = 10(5x - 2) \]\[ 3x - 1 = 50x - 20 \].Rearrange the terms to form:\[ 50x - 3x = 20 - 1 \].Simplify this to:\[ 47x = 21 \].

Step 4: Isolate x

Solve for \( x \) by dividing both sides by 47:\[ x = \frac{21}{47} \].

Step 5: Check the Solution

Substitute \( x = \frac{21}{47} \) back into the original logarithmic equations to ensure validity. Both expressions inside the logarithms must be positive:- \( 3 \times \frac{21}{47} -1 = \frac{63}{47} - 1 = \frac{16}{47} \) is positive.- \( 5 \times \frac{21}{47} -2 = \frac{105}{47} - 2 = \frac{11}{47} \) is positive. Thus the solution does not violate the domain conditions of logarithms.

Key Concepts

Logarithm Properties

A logarithm is a mathematical tool that helps us to work with exponential relationships by transforming multiplication into addition. There are several key properties of logarithms that make them very useful in solving equations. One important property is the quotient rule: \( \log a - \log b = \log \left( \frac{a}{b} \right) \). This property allows us to condense or expand logarithmic expressions based on their arithmetic operations.
These properties are critical when working with logarithmic equations because they enable us to simplify complex expressions. By applying these rules, such as in our case subtracting \( \log(5x-2) \) from \( \log(3x-1) \), we can rewrite the logarithmic equation into a simpler form for easier handling. As we saw, this approach holds the equation's balance and is instrumental in progressing toward a solution.

Solving Algebraic Equations

After applying logarithm properties, you often find yourself facing an algebraic equation. The goal at this stage is to solve for the unknown variable, usually denoted by \( x \). Solving algebraic equations involves a systematic approach often consisting of isolation and simplification of the variable terms.
In this problem, once the logarithms were condensed to a single equation, transforming it required clear algebraic manipulation. Initially, after rewriting the problem as \( \frac{3x-1}{5x-2} = 10 \), we needed to eliminate the fraction to allow a simpler algebraic equation. This step involved multiplying both sides of the equation by the denominator \( 5x-2 \), collapsing the fraction, and generating a linear equation: \( 3x - 1 = 50x - 20 \).
Rearranging the terms to bring all like terms to one side, and combining terms gives us the final form: \( 47x = 21 \), which is straightforward to solve by dividing both sides by 47.

Exponentiation

Exponentiation is the process of raising a number to a power and is fundamental when reversing logarithms. When you encounter an equation like \( \log_a(b) = c \), exponentiation is used to revert the logarithmic form into its exponential counterpart. The rule here is that \( a^c = b \), essentially stating that the base \( a \) raised to the power \( c \) equals \( b \).
In our logarithmic problem, we used the common base of 10 (since no base was specified, it defaults to 10) and applied exponentiation to both sides, transforming \( \log\left( \frac{3x-1}{5x-2} \right) = 1 \) into an expression where \( 10^1 = 10 \). This step transformed the logarithmic expression to a fraction \( \frac{3x-1}{5x-2} = 10 \), setting the stage for solving the algebraic equation as shown in earlier sections.

Checking Solutions

After solving an equation, it's crucial to verify that the solution is valid. Particularly with logarithms, you must ensure the solution doesn't lead to a logarithm of a non-positive number, as logarithms of zero or negative numbers are undefined in real numbers.
For our solution \( x = \frac{21}{47} \), checking involves substituting back into the original logarithmic expressions \( 3x-1 \) and \( 5x-2 \) and ensuring both evaluations result in positive numbers. Specifically, computing \( 3 \times \frac{21}{47} - 1 = \frac{16}{47} \) and \( 5 \times \frac{21}{47} - 2 = \frac{11}{47} \) reveals they are both positive, conforming to the domain restrictions on logarithms. This check validates that \( x = \frac{21}{47} \) is indeed a legitimate and correct solution.