Problem 88 - Introductory and Intermediate Algebra for College Students 4th

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. $$\log (x+3)+\log (x-2)=\log 14$$

Step-by-Step Solution

Answer

Therefore, the solution to the given logarithmic equation is \(x = 4\). Rejecting \(x = -5\) as it does not belong to the domain of the original logarithmic expressions.

Step 1: Combine the logs on the left side

Using the rule \(\log a + \log b = \log(ab)\), which allows one to combine two logs with the same base into a single log, the equation can be rewritten as \(\log ((x+3)(x-2)) = \log 14\)

Step 2: Set the arguments equal to each other

Since logarithms with the same base and equal values are equal, the arguments can be set equal to each other: \((x+3)(x-2)=14\). This results in a quadratic equation \(x^2 + x - 20 = 0\).

Step 3: Solve the quadratic equation

The quadratic formula or factoring can be used to solve the equation. By Factoring the equation it can be written as: \((x-4)(x+5) = 0\), which gives two solutions: \(x=4\) and \(x=-5\)

Step 4: Reject invalid solutions

Substituting \(x=4\) into the original equation: \(\log 7 + \log 2 = \log 14\), which is a true statement, so \(x = 4\) is a solution. However, substituting \(x=-5\) into the original equation results in undefined expressions \(\log -2\) and \(\log -7\), so \(x = -5\) is rejected and not a solution to the original logarithmic equation.