Problem 69 - 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. $$\log _{3}(x+6)+\log _{3}(x+4)=1$$

Step-by-Step Solution

Answer

The exact solution to the logarithmic equation is \(x = -3\).

Step 1: Combine the logs

Use the product rule of logarithms, \(\log_b{M} + \log_b{N} = \log_b{MN}\), to combine the two logs: \(\log_3{(x+6)(x+4)} = 1\)

Step 2: Remove the log

To remove the log, rewrite the equation in exponential form using the definition of logarithm, \(\log_b{A} = x \Rightarrow b^x = A \): \(3^1 = (x+6)(x+4)\)

Step 3: Expand and solve the equation

Expand the equation, \(3 = x^2 + 10x + 24\). Then subtract 3 from both sides to have a equation equals to zero: \(x^2 + 10x + 21 = 0\)

Step 4: Factor the equation

We can factor the quadratic equation to find the values of x: \((x+3)(x+7) = 0\). From here, solve the equation to get \(x = -3\) or \(x = -7\).

Step 5: Check the solutions

The original logarithmic equations only hold true for x values that are greater than -6 and -4. The solution \(x = -7\) is not in the domain of the original equation, so it must be rejected. The solution \(x = -3\) is in the domain and thus is the solution to the equation.