Problem 83 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log (7 y+1)=2 \log (y+3)-\log 2 $$

Step-by-Step Solution

Answer

The exact solutions are \( y = 7 \) and \( y = 1 \).

Step 1: Apply Logarithmic Properties

Start by applying the logarithm properties: \( \log a - \log b = \log \frac{a}{b} \) and \( n\log a = \log a^n \). This will help simplify the equation. The given equation is \( \log (7y+1) = 2 \log (y+3) - \log 2 \). Rewrite the terms on the right by applying these properties: \( 2 \log (y+3) = \log (y+3)^2 \) and \( \log (y+3)^2 - \log 2 = \log \frac{(y+3)^2}{2} \). The equation then becomes \( \log (7y+1) = \log \frac{(y+3)^2}{2} \).

Step 2: Set the Arguments Equal

Since the logarithms are equal, their arguments must be equal as well. Therefore, set the arguments inside the logarithms equal to each other: \( 7y + 1 = \frac{(y+3)^2}{2} \).

Step 3: Clear the Fraction

To eliminate the fraction, multiply both sides of the equation by 2: \( 2(7y + 1) = (y+3)^2 \). This simplifies to \( 14y + 2 = (y+3)^2 \).

Step 4: Expand and Simplify

Expand the squared term on the right: \((y+3)^2 = y^2 + 6y + 9\). Insert this back into the equation: \( 14y + 2 = y^2 + 6y + 9 \). Simplify by reorganizing into the standard form of a quadratic equation: \( 0 = y^2 - 8y + 7 \).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \( y^2 - 8y + 7 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -8 \), and \( c = 7 \). Calculate the discriminant: \( b^2 - 4ac = (-8)^2 - 4 \times 1 \times 7 = 64 - 28 = 36 \). Compute the roots: \( y = \frac{8 \pm \sqrt{36}}{2} \). Solve for \( y \): \( y = \frac{8 \pm 6}{2} \). This gives us the solutions \( y = 7 \) or \( y = 1 \).

Step 6: Verify Solutions and Check Conditions

Verify the solutions by substituting them back into the original equation to ensure they don't result in taking a logarithm of a non-positive number. Both \( y = 7 \) and \( y = 1 \) satisfy the original equation conditions as they result in positive arguments for the logarithms.

Key Concepts

logarithm properties

Logarithm properties are essential tools in solving logarithmic equations. One of the primary properties is the **Product Property**, which states that the logarithm of a product is the sum of the logarithms:

• \( \log(AB) = \log A + \log B \)

Another key property is the **Quotient Property**, used when subtracting one logarithm from another to express it as a division:

• \( \log(\frac{A}{B}) = \log A - \log B \)

Additionally, the **Power Property** expresses the logarithm of a power as a multiplication:

• \( n\log A = \log(A^n) \)

These properties allow you to manipulate and simplify complex logarithmic equations by transforming them into more manageable forms. In the given problem, they help turn a challenging equation into a solvable form by setting the arguments of the logs equal to one another.

quadratic equation

A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). It is characterized by the term \( x^2 \), where \( a \), \( b \), and \( c \) are coefficients, and \( a eq 0 \). Quadratic equations can appear in various forms, from standard to vertex forms, but the primary focus is on its **standard form**.
To solve a quadratic equation, several methods are available, such as factoring, using the quadratic formula, or completing the square. In the solution of our logarithmic equation, after applying logarithmic properties and simplifying, we end with the quadratic equation \( y^2 - 8y + 7 = 0 \).
Recognizing and solving quadratic equations is a fundamental skill as it appears in many branches of mathematics and real-world applications.

discriminant

The discriminant is a component of the quadratic formula, given by \( b^2 - 4ac \). It determines the nature and number of solutions of a quadratic equation.
Here's how the discriminant influences outcomes:

• If the discriminant is positive (\( b^2 - 4ac > 0 \)), the quadratic equation has two distinct real solutions.
• If it is zero (\( b^2 - 4ac = 0 \)), there is exactly one real solution, also known as a repeated or double root.
• If negative (\( b^2 - 4ac < 0 \)), the solutions are complex and not real numbers.

In our problem, calculating the discriminant: \( (-8)^2 - 4 \times 1 \times 7 = 36 \), showed it to be positive, thus indicating there are two distinct real solutions for the quadratic equation derived from the original logarithmic equation.

quadratic formula

The quadratic formula is a universal solution method for quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides the solutions as follows:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula is derived from completing the square on the general quadratic equation, and it applies to any quadratic equation where \( a eq 0 \).
In the given problem, after simplifying the original logarithmic equation to \( y^2 - 8y + 7 = 0 \), we used the quadratic formula with \( a = 1 \), \( b = -8 \), and \( c = 7 \) to find the values of \( y \).
Solving with the formula, we first consider the discriminant, which we found was \( 36 \). Thus, the solutions were computed as \( y = \frac{8 \pm 6}{2} \), giving us the exact roots \( y = 7 \) and \( y = 1 \). These roots were verified against conditions required by the logarithmic original expression, confirming them as valid solutions.