Problem 99 - Algebra for College Students

Question

Solve each equation. $$ \log _{10}(2 x-1)-\log _{10}(x-2)=1 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = \frac{19}{8}\).

Step 1: Use the Properties of Logarithms

Let's start by using the logarithmic property \(\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})\). Here, we have two logarithms with the same base, so we can combine them. Thus, \( \log_{10}(2x-1) - \log_{10}(x-2) \) becomes \( \log_{10}\left(\frac{2x-1}{x-2}\right) \).

Step 2: Set the Logarithmic Equation

Now that the equation is simplified using properties of logarithms, we have: \( \log_{10}\left(\frac{2x-1}{x-2}\right) = 1 \). To solve this, proceed to convert the log equation into its exponential form.

Step 3: Convert to Exponential Form

Any equation of the form \(\log_b(a) = c\) can be rewritten in exponential form as \(a = b^c\). For our equation, this means \( \frac{2x-1}{x-2} = 10^1 \). Hence, \( \frac{2x-1}{x-2} = 10 \).

Step 4: Solve the Equation for x

We now solve the equation \(\frac{2x-1}{x-2} = 10\). Start by cross-multiplying: \(2x-1 = 10(x-2)\). Expand and rearrange to get \(2x-1 = 10x-20\). Simplify this to find \(8x = 19\), giving us \(x = \frac{19}{8}\).

Step 5: Verify the Solution

Verify that this value of \(x\), which is \(\frac{19}{8}\), makes sense in the original equation. Substitute back to check if \(\log_{10}(2x-1)\) and \(\log_{10}(x-2)\) result in real values. Calculate \(2x-1 = \frac{29}{8}\) and \(x-2 = \frac{3}{8}\). Both are positive, so the logarithms are defined.

Key Concepts

Properties of Logarithms

Logarithms have useful properties that can simplify complex equations, making them easier to solve. One key property in our exercise is the difference of two logarithms, which can be combined into a single logarithm. This property is expressed as:

• \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \)

This is extremely useful when both logarithms have the same base, as they do in this exercise (base 10). By applying this rule, we take two separate logarithms and transform them into a single one, which simplifies the equation significantly.
This simplification allows us to focus on comparing expressions inside a single logarithmic function rather than dealing with multiple logarithmic terms.

Exponential Form

After simplifying the equation using logarithmic properties, we encounter another key concept: converting a logarithmic equation to an exponential form. This conversion is straightforward and very effective for solving logarithmic equations.
A logarithmic equation \( \log_b(a) = c \) can be rewritten as:

• \( a = b^c \)

In this exercise, the equation \( \log_{10}\left(\frac{2x-1}{x-2}\right) = 1 \) is rewritten as \( \frac{2x-1}{x-2} = 10^1 \). Recognizing that 10 to the power of 1 is simply 10, the equation becomes \( \frac{2x-1}{x-2} = 10 \). This transformation turns a logarithmic problem into a more familiar algebraic form that can easily be manipulated.

Cross-Multiplication

Once the equation is in an algebraic form, we can solve it by cross-multiplication. Cross-multiplication is a technique used to remove fractions from equations by multiplying across from numerator to denominator.
For example, to solve \(\frac{2x-1}{x-2} = 10\), you cross-multiply to eliminate the fraction:

• Multiply \(2x-1\) by 1, and \(x-2\) by 10.

This results in the equation \(2x-1 = 10(x-2)\), which is simplified further by expanding and rearranging to isolate \(x\). Using algebraic manipulation, we simplify this to \( 8x = 19 \), and solve for \(x\) by dividing both sides by 8, finding \( x = \frac{19}{8} \).
Cross-multiplication is a powerful and straightforward method for clearing fractions from an equation, leading to an easier solution path.

Verifying Solutions

Verification of solutions is a crucial final step in solving any equation, including logarithmic ones. It ensures that the solution is valid within the context of the original equation and that it does not result in any undefined mathematical operations, like the logarithm of a negative number.
In this exercise, you verify the solution by substituting \( x = \frac{19}{8} \) back into the original logarithmic expressions:

• Calculate \(2x-1\) to ensure it's greater than zero.
• Calculate \(x-2\) to ensure it's greater than zero.

Both calculations must result in positive values for the logarithms to be defined, as logarithms cannot process negative numbers. Here,

• \(2x-1 = \frac{29}{8} > 0\), and
• \(x-2 = \frac{3}{8} > 0\).

Since both conditions are satisfied, the solution is verified. Verification ensures confidence in the solution's accuracy and relevance, confirming that all mathematical principles are honored.