Problem 88 - 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. $$ \log (x+3)+\log (x-2)=\log 14 $$

Step-by-Step Solution

Answer

The solution is \(x = 4\) since it is the only valid solution within the domain of the original logarithmic expressions.

Step 1: Applying the logarithm rule

The logarithm law mentioned above is applied, which gives \(\log[(x+3)*(x-2)] = \log(14)\).

Step 2: Solving for x

Next, since logarithms with same bases are equal then the arguments should be equal too. So, the equation becomes \((x+3)*(x-2) = 14\). Set this quadratic equation to zero to solve for x, so the equation becomes \((x+3)*(x-2) - 14 = 0\). This simplifies to \(x^2 + x -20 = 0\).

Step 3: Factoring the quadratic equation

Factoring the left hand side, we get \((x+5)(x-4) = 0\). Setting each factor equal to zero and solving for x, we get two possible solutions: \(x = -5\) and \(x = 4\).

Step 4: Verifying the solutions are within domain

The original equation has \(\log (x+3)\) and \(\log (x-2)\). Given this, \(x\) in \(\log (x+3)\) should be >-3 and \(x\) in \(\log (x-2)\) should be >2. Thus, between the two solutions we obtained (\(-5\) and \(4\)), only \(4\) is contained in both domains, so the solution for \(x\) is \(4\).

Key Concepts

Logarithm Rules

Understanding logarithm rules is crucial when solving logarithmic equations. Logarithms are the inverse operations of exponentiation. The basic logarithmic function is expressed as \(\log_b(x) = y\) if and only if \(b^y = x\). Here, \(b\) is the base, \(x\) is the argument, and \(y\) is the logarithm of \(x\) to the base \(b\). When solving logarithmic equations, there are several rules that can make the process easier.

• The Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\), which allows the addition of logarithms to be expressed as the logarithm of a product.
• The Quotient Rule: \(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\), useful for rewriting the subtraction of logarithms as a single logarithm of a quotient.
• The Power Rule: \(\log_b(M^k) = k\log_b(M)\), which says that the logarithm of a power can be rewritten as the exponent times the logarithm of the base.

These rules can transform complex logarithmic expressions into simpler ones that are more easily solvable.

Quadratic Equations

A quadratic equation is a second-order polynomial equation in a single variable \(x\) with a non-zero coefficient for \(x^2\). It has the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. To solve a quadratic equation, one may use factoring, completing the square, using the quadratic formula, or graphing.

For instance, if we consider the equation \(x^2 + x - 20 = 0\) derived from our logarithmic equation above, it represents a quadratic equation. The solutions of this equation, also known as roots or zeros, can be found by various methods, including the process of factoring when the quadratic is factorable.

Factoring Quadratics

Factoring quadratics is a process of breaking down the quadratic equation into a product of binomials. Essentially, you're looking for two numbers that add up to the coefficient of \(x\) and multiply to the constant term. This can often simplify the process of finding the roots.

For example, the quadratic \(x^2 + x - 20 = 0\) factors into \(x+5)(x-4) = 0\). The factored form makes it evident that the roots are \(x = -5\) and \(x = 4\). However, it’s essential to check that these solutions adhere to the constraints of the original problem, such as the domain of a logarithmic function.

Domain of Logarithmic Functions

The domain of logarithmic functions is the set of all permissible input values (\(x\)-values) that make the function 'work' and result in real number outputs. As a fundamental property, the argument of a logarithm must be positive, since the logarithm of zero or a negative number is undefined.

In the context of the given exercise, the functions \(\log (x+3)\) and \(\log (x-2)\) imply that \(x+3 > 0\) and \(x-2 > 0\), respectively. This translates to \(x > -3\) and \(x > 2\). When solving a logarithmic equation, always ensure to reject any solution not within the domain of the original logarithmic expressions. For the given problem, only the solution \(x = 4\) is valid, as it satisfies the domain requirements of both logarithms involved in the equation.