Problem 84 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 2 \log (y+2)=\log (y+2)-\log 12 $$

Step-by-Step Solution

Answer

The valid solution is \(y = -\frac{23}{12} \approx -1.9167\).

Step 1: Equate the Logarithmic Expressions

Using the properties of logarithms, recall that if \( ext{log}_b(M) = ext{log}_b(N)\), then \(M = N\). Here we have: \ \[2 \log (y+2) = \log (y+2) - \log 12.\] \ Let's simplify these expressions to solve \(y\).

Step 2: Simplify the Left-hand Side

The equation \(2 \log(y+2)\) can be rewritten using the power rule of logarithms: \ \[2 \log(y+2) = \log((y+2)^2).\] \ Now our equation is: \ \[\log((y+2)^2) = \log(y+2) - \log 12.\]

Step 3: Simplify the Right-hand Side

Apply the quotient rule of logarithms: \(\log(A) - \log(B) = \log(A/B)\). Therefore, \ \[\log(y+2) - \log(12) = \log \left(\frac{y+2}{12}\right).\] \ The equation becomes: \ \[\log((y+2)^2) = \log \left(\frac{y+2}{12}\right).\]

Step 4: Set the Arguments Equal

Since the logarithms are equal, set the arguments equal: \ \[(y+2)^2 = \frac{y+2}{12}.\] \ Multiply both sides by 12 to eliminate the fraction: \ \[12(y+2)^2 = y+2.\]

Step 5: Expand and Collect Like Terms

Expand \(12(y+2)^2\) and set the equation to zero. This gives: \ \[12(y+2)^2 = (y+2) \rightarrow 12(y^2+4y+4) = y+2.\] \ \[12y^2 + 48y + 48 = y+2.\] \ Rearrange all terms to one side: \ \[12y^2 + 47y + 46 = 0.\]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(y = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\), where \(a=12\), \(b=47\), and \(c=46\). Calculate the discriminant: \ \[b^2 - 4ac = (47)^2 - 4(12)(46) = 2209 - 2208 = 1.\] \ Now substitute in the quadratic formula: \ \[y = \frac{{-47 \pm \sqrt{1}}}{{24}}.\] \ \[y = \frac{{-47 \pm 1}}{{24}}.\] \ Thus, \[y = -2\] or \[y = -\frac{23}{12}.\]

Step 7: Verify the Solutions in the Context of Logarithms

For a logarithm to be valid, the argument must be greater than zero. Thus, check the candidates: \ For \(y=-2\), the argument \((y+2) = 0\) is not valid as \(\log(0)\) is undefined. \ For \(y=-\frac{23}{12}\), the argument \((y+2) = \frac{1}{12}\) is valid as it is greater than zero.

Key Concepts

Quadratic Formula

One of the fundamental tools in algebra is the quadratic formula. We use it to find the solutions, or roots, of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The quadratic formula is: \[y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]Here, \( a \), \( b \), and \( c \) are constants from the equation, and \( b^2-4ac \) is called the discriminant.

• If the discriminant is positive, we have two real and distinct solutions.
• If it's zero, we get one real solution (a repeated root).
• If the discriminant is negative, the solutions are complex and not real numbers.

In our exercise, the quadratic equation derived from the log equation was \( 12y^2 + 47y + 46 = 0 \). By calculating the discriminant, \(b^2 - 4ac = 1\), we found it to be positive, meaning we have two distinct solutions. Using the quadratic formula, the solutions were computed as \( y = -2 \) and \( y = -\frac{23}{12} \). Checking the validity within the context is crucial, especially when dealing with logarithms.

Logarithm Properties

Logarithmic properties are powerful tools to simplify and solve equations involving logarithms. These properties include the product rule, quotient rule, and power rule:

• Product Rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Rule: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Rule: \( \log_b(M^k) = k \cdot \log_b(M) \)

In the given exercise, the quotient and power rules were necessary to simplify the equation:
Originally, the equation \( 2 \log(y+2) = \log(y+2) - \log(12) \) was simplified by using the power rule, changing it to \( \log((y+2)^2) \). Then, the quotient rule transformed \( \log(y+2) - \log(12) \) into \( \log\left(\frac{y+2}{12}\right) \). By equalizing the arguments of the logarithms, we set \( (y+2)^2 = \frac{y+2}{12} \), thus transitioning to a quadratic equation.

Quadratic Equations

A quadratic equation is any equation that can be restructured into the standard form \( ax^2 + bx + c = 0 \). Key elements of solving these equations involve methods like factorization, completing the square, or using the quadratic formula.
In our problem, we derived a quadratic equation from a logarithmic equation. Once we had \( (y+2)^2 = \frac{y+2}{12} \), we multiplied through by 12 to eliminate the fraction, leading to \( 12(y+2)^2 = y+2 \).
Expanding \( 12(y+2)^2 \) results in a new equation: \( 12y^2 + 48y + 48 = y+2 \).

• Organize all terms to one side to achieve the quadratic form: \( 12y^2 + 47y + 46 = 0 \).
• Use the quadratic formula, the most versatile method, to find the potential solutions.

Verifying solutions is critical when derived from other forms, like logarithms, to ensure they are valid in the original equation's context.