Question

The solution set of the equation \(\log (x+6)-\log 8=\log 9-\log (x+7)\) is (1) \(\\{-15,2\\}\) (2) \(\\{2\\}\) (3) \(\\{-15,0,2\\}\) (4) \(\\{0,2\\}\)

Short answer

Answer: (1) \(\\{-15,2\\}\)

Step-by-step solution

Use the logarithmic property: \(\log a-\log b=\log \frac{a}{b}\) to rewrite the given equation as: \(\log \frac{x+6}{8} = \log \frac{9}{x+7}\) #Step 2: Remove the logarithms#

Since both sides of the equation have logarithms, we can remove the logarithms by setting the arguments equal to each other: \(\frac{x+6}{8}=\frac{9}{x+7}\) #Step 3: Cross-multiply the equation#

To solve for x, we'll cross-multiply the equation: \((x+6)(x+7)=8 \cdot 9\) #Step 4: Expand and simplify the equation#

Expand the left side of the equation and simplify the right side: \(x^2+7x+6x+42=72\) Now, combine like terms, and subtract 72 from both sides: \(x^2+13x-30=0\) #Step 5: Factor the quadratic equation#

To solve for x, factor the quadratic equation: \((x+15)(x-2)=0\) #Step 6: Solve for x#

Now, set each factor equal to zero and solve for x: \(x+15=0 \Rightarrow x=-15\) \(x-2=0 \Rightarrow x=2\) So, the solution set of the equation is: \(\\{-15,2\\}\) The correct answer is (1) \(\\{-15,2\\}\).

Key concepts

Quadratic Equations

Quadratic equations are a very common part of algebra, and they appear in many kinds of problems. They have the standard form of \[ax^2 + bx + c = 0,\]where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). In our exercise, after handling the logarithm equation, a quadratic equation emerged: \[x^2+13x-30=0.\]This equation is derived from cross-multiplying the fractions resulting from the logarithmic terms and then simplifying. Understanding how to work with quadratic equations is crucial because they have specific solutions that can often be found by factoring, using the quadratic formula, or completing the square.

Most students find factoring the simplest approach when possible, as it can directly give the roots of the equation. In our case, once the equation was in standard form, it was straightforward to factor it into \((x+15)(x-2)=0.\)

Algebraic Manipulation

Algebraic manipulation is a skill that involves rearranging equations to isolate and solve for a particular variable. It often involves using principles such as balancing, distributing, factoring, and applying laws such as the properties of logarithms.

In our problem, we started with a logarithmic equation \[\log \frac{x+6}{8} = \log \frac{9}{x+7}.\]Using the property that \(\log a = \log b\) implies \(a = b\), we eliminated the logarithms, leading us to\[ \frac{x+6}{8} = \frac{9}{x+7}. \]From this point, cross-multiplying helped us eliminate the fractions and gave an equation that was easier to manipulate and solve. Algebraic manipulation helped transform a complex logarithmic equation into a manageable quadratic form.

Solution Sets

The solution set for an equation represents all the possible values of the variable that satisfy the equation. By solving the quadratic equation from our problem, \[(x+15)(x-2)=0,\]we found two potential solutions: \(x = -15\) and \(x = 2\).

However, it's crucial to verify these values with the original equation to ensure they are valid in the given context. This is particularly important for logarithmic equations, as they can impose certain restrictions on the domain of possible solutions. Typically for logarithmic expressions, the arguments must be positive, so substitutions into the original equation determine whether all supposed solutions are permissible.

After verification, \(x = -15\) and \(x = 2\) satisfy the equation, giving us the final solution set: \{-15, 2\}.