Problem 70 - College Algebra Essentials

Question

Solve each logarithmic equation. 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 for the given logarithmic equation is \(x = -1\).

Step 1: Using the properties of logarithms

Use the property of logs which states that the sum of the logs equals the log of the product of their numbers to combine \(\log _{6}(x+3)\) and \(\log _{6}(x+4)\). This results in \( \log _{6}[(x+3) * (x+4)] = 1\)

Step 2: Converting the Logarithmic Equation into an Exponential Equation

To convert the logarithmic equation into an exponential equation, remember that \(\log _{b}(a) = c\) is equivalent to \(b^c = a\). Therefore, \(\log _{6}[(x+3) * (x+4)] = 1\) can be rewritten as \(6^1 = (x+3)*(x+4)\).

Step 3: Solving for \(x\)

Now that it has been simplified to \(6 = (x+3)*(x+4)\), expand the right-hand side and simplify further to obtain a quadratic equation: \(x^2 + 7x + 12 - 6 = 0\), which simplifies to: \(x^2 + 7x + 6 = 0\). This can be factored to \((x + 1)(x + 6) = 0.\) Solving for \(x\), we have \(x = -1, -6.\)

Step 4: Checking for extraneous solutions

We must double-check if these solutions are permissible by checking if they are in the domain of the original log expressions. The inputs to a logarithm must be greater than zero. Using \(x = -1,\) you get inputs of \(2\) and \(3\) for the original logs, which are valid. However, using \(x = -6,\) you get inputs of \(-3\) and \(-2\) which are not valid as they are not greater than zero. So, \(x = -6\) is an extraneous solution and must be rejected. Therefore, the solution to the equation is \(x = -1.\)