Problem 83 - 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 \(y = 1\); both satisfy the equation.

Step 1: Use Properties of Logarithms - Condense the Right Side

Recall the property of logarithms that states \(a \log b = \log b^a\). Apply this to the right side: \(2 \log (y+3) = \log((y+3)^2)\). Now, use the subtraction property of logarithms, \(\log a - \log b = \log\left(\frac{a}{b}\right)\), to rewrite:\(\log((y+3)^2) - \log 2 = \log\left(\frac{(y+3)^2}{2}\right)\). Thus, the equation now is \(\log(7y + 1) = \log\left(\frac{(y+3)^2}{2}\right)\).

Step 2: Eliminate Logarithms - Equate the Arguments

Since the logarithms on both sides are equal, their arguments must also be equal. Hence:\(7y + 1 = \frac{(y+3)^2}{2}\).

Step 3: Clear the Fraction - Multiply Both Sides by 2

To eliminate the fraction, multiply both sides of the equation by 2:\(2(7y + 1) = (y + 3)^2\).Simplifying the left side gives: \(14y + 2 = (y+3)^2\).

Step 4: Expand and Rearrange - Form a Quadratic Equation

Expand \((y+3)^2\): \((y + 3)^2 = y^2 + 6y + 9\).Thus the equation becomes\(14y + 2 = y^2 + 6y + 9\).Reorder to form a quadratic equation:\(y^2 + 6y + 9 - 14y - 2 = 0\).Simplify to:\(y^2 - 8y + 7 = 0\).

Step 5: Solve the Quadratic Equation - Factoring

Factor the quadratic equation:\(y^2 - 8y + 7 = 0\) can be factored as \((y - 7)(y - 1) = 0\).Thus, the solutions are \(y = 7\) and \(y = 1\).

Step 6: Verify the Solutions - Check Within Original Equation

Substitute \(y = 7\) into the original logarithmic equation to check:\(\log(7 \times 7 + 1) = \log(50)\), and check \(2 \log(7 + 3) - \log 2 = \log\left(\frac{100}{2}\right) = \log(50)\).Substitute \(y = 1\):\(\log(7 \times 1 + 1) = \log(8)\), and check \(2 \log(1 + 3) - \log 2 = \log\left(\frac{16}{2}\right) = \log(8)\).Both solutions verify.

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is a crucial step in solving logarithmic equations effectively. In our exercise, these properties helped us simplify and manipulate the equation.One property we used is the power rule, which states that multiplying a logarithm by a number is equivalent to the logarithm of the power of the number. Formally, it's expressed as:

• \(a \log b = \log b^a\)

Another important property is the subtraction rule, which allows us to simplify the difference of two logarithms to a single logarithm. It's expressed as:

• \(\log a - \log b = \log\left(\frac{a}{b}\right)\)

By using these properties, the original equation was condensed from multiple logarithmic expressions on the right side to a single expression, making it easier to work with.

Quadratic Equation

After manipulating the logarithmic equation, we ended up with a quadratic equation. A quadratic equation is a second-degree polynomial equation of the form:

• \(ax^2 + bx + c = 0\)

In our solved exercise, the quadratic equation took the form:

• \(y^2 - 8y + 7 = 0\)

These kinds of equations often have two solutions. They are typically solved by factoring, using the quadratic formula, or by completing the square. Understanding this concept is essential because it's a common form of an equation you will encounter in mathematics.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as a product of two binomials. In essence, an equation like \(y^2 - 8y + 7 = 0\) can be rewritten using factoring as:

• \((y - 7)(y - 1) = 0\)

Through factoring, you can easily find the solutions to the quadratic equation. The factors provide the values of \(y\) where each binomial equals zero:

• If \((y - 7) = 0\), then \(y = 7\)
• If \((y - 1) = 0\), then \(y = 1\)

These solutions arise because the product of the factors equals zero only when at least one of the factors is zero. Factoring is not just a tool for solving equations but also helps in simplifying expressions in algebra.

Verifying Solutions

Once we solve an equation, especially a transformed one, verifying the solutions is crucial to ensure they satisfy the original equation.In this exercise, after finding the possible solutions \(y = 7\) and \(y = 1\), we substitute them back into the original logarithmic equation. This confirms their validity:

• For \(y = 7\), substituting gives \(\log(50) = \log(50)\)
• For \(y = 1\), substituting gives \(\log(8) = \log(8)\)

Both substitutions hold true, confirming both solutions are correct. Verification helps eliminate errors that might arise during manipulation and ensures that the solutions conform to the original problem.