Problem 17 - Algebra 2

Question

Solve each inequality. Check your solutions. \(\log _{3}(2 x-1) \leq 2\)

Step-by-Step Solution

Answer

\( \frac{1}{2} < x \leq 5 \)

Step 1: Understand the Inequality

The given inequality is \(\log_{3}(2x-1) \leq 2\). This means we need to determine the values of \(x\) for which the logarithm base 3 of \(2x-1\) is less than or equal to 2.

Step 2: Rewrite the Inequality in Exponential Form

To eliminate the logarithm, we convert the inequality from logarithmic to exponential form. Since \(\log_{3}(2x-1) \leq 2\), this is equivalent to saying that \(2x-1 \leq 3^{2}\).

Step 3: Solve the Exponential Inequality

Simplify the inequality \(2x-1 \leq 9\) (since \(3^{2} = 9\)). Add 1 to both sides to get \(2x \leq 10\). Then, divide both sides by 2 to find the solution for \(x\), yielding \(x \leq 5\).

Step 4: Identify the Domain of the Logarithmic Function

Logarithmic functions are defined only for positive arguments. Therefore, \(2x-1 > 0\), which simplifies to \(x > \frac{1}{2}\).

Step 5: Find the Intersection of Solutions

The solution to the inequality must satisfy both conditions: \(x \leq 5\) and \(x > \frac{1}{2}\). The intersection of these solutions is \(\frac{1}{2} < x \leq 5\).

Step 6: Check the Solutions

Ensure the values in the interval \(\frac{1}{2} < x \leq 5\) satisfy the original inequality. For example, test \(x = 1\) which gives \(\log_{3}(1) = 0 \leq 2\) and \(x = 5\) which gives \(\log_{3}(9) = 2\), verifying the solution set is correct.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a variable appears as an exponent. They are defined as functions of the form \( f(x) = a^x \), where \( a \) is a constant called the base and \( x \) is any real number. The base \( a \) is usually a positive number greater than zero and not equal to one.In the context of solving inequalities involving logarithms, exponential functions come into play when we need to convert a logarithmic inequality into a form that is easier to manipulate. For example, if we have a logarithmic equation \( \log_{b}(y) = c \), we can rewrite this in exponential form as \( y = b^c \). In our original problem, this conversion helps solve the inequality \( \log_{3}(2x-1) \leq 2 \) by turning it into an exponential inequality \( 2x-1 \leq 9 \). This allows us to manage the inequality using basic algebraic techniques.Understanding exponential functions is crucial as they model various real-world phenomena such as population growth, radioactive decay, and interest compounded in financial calculations. In mathematics, they serve as an inverse to logarithmic functions, adding a layer of flexibility to solve complex equations and inequalities.

Inequality Solving

Solving inequalities is a fundamental skill in mathematics that involves finding the set of values that satisfy a given inequality condition. Inequality solving requires one to use similar skills as those used in solving equations but with careful attention to the inequality signs.When dealing with the inequality \( \log_{3}(2x-1) \leq 2 \), the first step is to eliminate the logarithmic format by converting it into an exponential form, which becomes \( 2x-1 \leq 9 \). Solving this inequality involves simple algebra. We add 1 to both sides, obtaining \( 2x \leq 10 \). Then, by dividing both sides by 2, we arrive at \( x \leq 5 \). This result indicates all values less than or equal to 5 are solutions to this transformed inequality.However, we also need to consider the condition that the expression inside the logarithm is positive, leading to another inequality \( 2x-1 > 0 \), simplifying to \( x > \frac{1}{2} \). The solution to the original problem must satisfy both conditions: \( x \leq 5 \) and \( x > \frac{1}{2} \). By intersecting these solutions, we find \( \frac{1}{2} < x \leq 5 \). This is the interval where all the values of \( x \) meet the original condition of the inequality.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They are generally expressed as \( f(x) = \log_{b}x \), where \( b \) is the base of the logarithm, and \( x \) is its argument. The logarithm \( \log_{b}x \) answers the question: "To what power must \( b \) be raised to obtain \( x \)?"In the given inequality \( \log_{3}(2x-1) \leq 2 \), the expression \( \log_{3}(2x-1) \) relies on the base 3. For logarithmic functions to be defined, the argument \( (2x-1) \) of the logarithm must always be positive. This stipulation ensures the function remains within its domain, requiring us to set the additional condition \( 2x-1 > 0 \) or \( x > \frac{1}{2} \).Logarithms simplify complex multiplicative sequences into additive ones, which is precisely what makes them powerful tools in both theoretical and applied contexts. They are particularly applicable in problems involving exponential growth or decay, as well as any mathematical problems needing simplification of powers and roots. By handling logarithms with care—particularly in inequalities—we ensure the underlying mathematical expressions retain their validity and applicability across different scenarios.