Problem 19 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{5}(3 x-1)=\log _{5}\left(2 x^{2}\right)\)

Step-by-Step Solution

Answer

The solutions are \( x = 1 \) and \( x = \frac{1}{2} \).

Step 1: Understand the property of logarithms

When two logarithms with the same base are equal, their arguments can be set equal to each other. That means if \( \log_{5}(3x - 1) = \log_{5}(2x^2) \), then \( 3x - 1 = 2x^2 \).

Step 2: Set the arguments equal

From the property discussed in Step 1, we write the equation without logarithms: \( 3x - 1 = 2x^2 \).

Step 3: Rearrange into a standard quadratic equation

Rearrange the equation so all terms are on one side, giving: \( 2x^2 - 3x + 1 = 0 \). This is now a quadratic equation.

Step 4: Apply the quadratic formula

To solve the quadratic equation \( 2x^2 - 3x + 1 = 0 \), we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = 1 \).

Step 5: Calculate the discriminant

First, calculate the discriminant: \( b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \). The discriminant is positive, so there are two real solutions.

Step 6: Solve for the roots using the quadratic formula

Using the quadratic formula with the discriminant: \( x = \frac{-(-3) \pm \sqrt{1}}{4} \), compute the solutions. This gives \( x = \frac{3 \pm 1}{4} \).

Step 7: Calculate the individual solutions

The two possible solutions are \( x = \frac{3 + 1}{4} = 1 \) and \( x = \frac{3 - 1}{4} = \frac{1}{2} \).

Step 8: Verify the solutions

Check each solution by substituting back into the original expressions for the arguments of the logarithms. For \( x = 1 \), both sides will calculate to \( 2 \). For \( x = \frac{1}{2} \), both sides will calculate to \( \frac{1}{2} \), verifying both solutions.

Key Concepts

Logarithmic Equations

Logarithmic equations involve equations where logarithms appear with the unknown variable in them. In our exercise, the equation given is \( \log _{5}(3x-1)=\log _{5}(2x^{2}) \). The key to solving such equations is to leverage the property that allows us to equate the arguments when the logarithms have the same base. This process considerably simplifies the equation.
In simpler terms, if you have two logs that are equal with identical bases, you can remove the logs and set their insides equal to one another. This means \( \log_b(M) = \log_b(N) \) leads to \( M = N \). It’s important to ensure that both sides of your resulting equation remain defined – in other words, the arguments of the logs must be positive for solutions to be valid.

Properties of Logarithms

Understanding the properties of logarithms is a fundamental skill when solving logarithmic equations. Here are some of the properties that are useful:

• The Product Rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \).
• The Quotient Rule: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \).
• The Power Rule: \( \log_b(M^n) = n \cdot \log_b(M) \).
• The Change of Base Formula: \( \log_b(M) = \frac{\log_k(M)}{\log_k(b)} \).

These rules aren't directly applied in our specific exercise but are valuable whenever dealing with more complex logarithmic problems. Recognizing these rules can help simplify expressions and solve equations effectively.

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides the solutions for \( x \). Here, \( a \), \( b \), and \( c \) are coefficients in the equation.
To apply it, it's vital first to rearrange your quadratic equation into its standard form, just like in our exercise with \( 2x^2 - 3x + 1 = 0 \). From there:

• Calculate the discriminant \( b^2 - 4ac \).
• Use the discriminant to find the number and nature of the roots.
• Substitute \( a \), \( b \), and \( c \) into the formula and solve for \( x \).

The formula is particularly useful for equations that do not easily factor and ensures that you can find solutions even if they aren’t integers.

Discriminant

The discriminant is an expression \( D = b^2 - 4ac \) derived from the quadratic formula and is very telling about the nature of the roots of a quadratic equation.
Here's what you can learn from it:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is exactly one real root, which is sometimes referred to as a repeated or double root.
• If \( D < 0 \), the roots are complex and occur as a conjugate pair.

This step is crucial for understanding what kind of solutions you’re dealing with after applying the quadratic formula. In our specific problem, the positive discriminant indicated two distinct solutions: \( x = 1 \) and \( x = \frac{1}{2} \). So, always calculate the discriminant to anticipate the type of solutions you'll encounter.