Problem 70 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{6}(x+3)+\log _{6}(x+4)=1 $$

Step-by-Step Solution

Answer

The solution to the given logarithmic equation is \(x = -1\).

Step 1: Use the property of logarithms

Apply the property of logarithms that states that the sum of two logarithms is equal to the logarithm of the products of their arguments. The equation now becomes: \[ \log _{6}[(x+3)(x+4)]=1.\]

Step 2: Solve for the argument

Convert the log equation to an exponential equation, then simplify to obtain a quadratic equation. Thus, \[(x+3)(x+4) = 6^1.\] This simplifies to \(x^2+7x+12=6\). Subtract 6 from both sides to set the equation to zero; obtaining \(x^2 + 7x + 6 = 0\).

Step 3: Factorize the quadratic equation

Factorize the quadratic equation to isolate \(x\), so it becomes \((x + 1)(x + 6) = 0\).

Step 4: Solve for \(x\)

Set each of the factors equal to zero and solve for \(x\), resulting in \(x + 1 = 0\) or \(x + 6 = 0\). Solving these gives \(x = -1\) or \(x = -6\)

Step 5: Check domain of original logarithmic expressions

The original logarithmic expressions had arguments \(x + 3\) and \(x + 4\). Logarithms are only defined for their arguments to be greater than zero. This excludes \(x = -6\) from our solution set, leaving us only with \(x = -1\).

Key Concepts

Properties of Logarithms

Logarithmic equations often leverage the properties of logarithms to simplify complex expressions. One of the most important properties is the addition rule for logarithms:

• \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \)
• This means that the sum of two logarithms with the same base is equal to a single logarithm of the product.

In the given problem, \( \log_6(x+3) + \log_6(x+4) = 1 \), we apply this rule to condense the equation into a single logarithmic expression: \[\log_6[(x+3)(x+4)] = 1\]This transformation is crucial as it simplifies solving the equation by reducing it into a form where the argument of the logarithm can be expressed and manipulated more easily. As we have rewritten the logarithm, the next step involves evaluating the logarithmic expression through conversion into an exponential form.

Quadratic Equations

Once the logarithmic equation has been simplified using properties of logarithms, the next step often involves dealing with a resulting algebraic expression like a quadratic equation. After converting the logarithmic equation to its exponential form based on the base of the logarithm, you should have:\[(x + 3)(x + 4) = 6\]Expanding this results in a quadratic equation: \[x^2 + 7x + 12 = 6\]To solve for \(x\), you need the equation to equal zero: \[x^2 + 7x + 6 = 0\]Quadratic equations can sometimes be solved by factoring. Factoring the quadratic helps isolate solutions for \(x\):\[(x + 1)(x + 6) = 0\]Setting each factor to zero provides potential solutions: \(x = -1\) or \(x = -6\). But, solving quadratic equations doesn’t end here; you must then check whether these solutions fit within the given constraints of the original problem.

Domain of Logarithmic Functions

In solving logarithmic equations, it's essential to consider the domain of the logarithmic functions involved. Logarithms are only defined for positive arguments. This means that any realistic solution for \(x\) needs to respect this constraint. For our original problem with expressions \(x + 3\) and \(x + 4\), we need:

• \( x + 3 > 0 \)
• \( x + 4 > 0 \)

This translates to:

• \( x > -3 \)
• \( x > -4 \)

Therefore, \(x\) must be greater than \(-3\), effectively eliminating any valid solutions which do not meet this requirement. When we solve for \(x\), potential solutions included \(x = -1\) and \(x = -6\), but \(x = -6\) does not satisfy \(x > -3\). Hence, the only suitable solution within the domain of the logarithmic function is \(x = -1\). This kind of checking ensures that all final answers are not only mathematically accurate but also valid under the conditions set by our logarithmic expressions.