Problem 76 - 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. $$ \log _{2}(x-3)+\log _{2} x-\log _{2}(x+2)=2 $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x = 8\).

Step 1: Combine logarithmic terms

According to the properties of logarithms, when multiple logarithmic expressions with the same base are added, they can be combined as a single expression with the product of the contents. In this case, the given equation \(\log _{2}(x-3)+\log _{2} x-\log _{2}(x+2)=2\) can be rewritten as \(\log _{2}((x-3) * x / (x+2)) = 2.\)

Step 2: Rewrite the equation in exponential form

Converting the logarithm to an exponential, the base is 2, the exponent is 2 and the result is \((x-3)x/(x+2)\). The original equation can be rewritten as: \(2^2 = (x-3) * x / (x+2)\). Simplify this to get \(4 = (x^2 - 3x) / (x+2)\).

Step 3: Multiply both sides of the equation by \(x+2\)

To remove the denominator, multiply the equation by \(x+2\). When doing so, remember to distribute the multiplication across the terms on the right side of the equation. This results in \(4(x+2) = x^2 - 3x\). After simplifying, this can be written as \(4x + 8 = x^2 - 3x.\)

Step 4: Rearrange the equation to form a quadratic equation

To solve for \(x\), rewrite the equation in the standard form of a quadratic equation as \(x^2 - 7x - 8 = 0\).

Step 5: Factor the quadratic equation

The equation can then be factored as \((x - 8)(x + 1) = 0\). According to the zero-product property, if the product of these factors equals zero, then either \(x - 8 = 0\) or \(x + 1 = 0\)

Step 6: Solve for \(x\)

By setting these factors equal to zero, two potential solutions for \(x\) can be found: \(x = 8\) and \(x = -1\). However, \(x = -1\) is not in the domain of the original logarithmic expressions. Therefore, \(x = -1\) must be rejected.

Step 7: Check solution is in the domain of the original logarithmic expressions

As mentioned, \(x = -1\) cannot be the solution because it would result in a negative number within a logarithm (specifically, for \(\log_{2}(x)\)). Therefore, only \(x = 8\) is an actual solution to the original equation.

Key Concepts

Properties of Logarithms

The properties of logarithms are foundational in solving logarithmic equations, like the one we see in the exercise. A logarithm, written as \(\text{log}_{b}(a)\), asks the question 'to what power must we raise \(b\) to obtain \(a\)?' This relationship is crucial because it ties together exponents and logarithms.

Several properties help us manipulate and combine logarithmic terms. One such property is the product rule, stating that the sum of two logarithms with the same base can be combined into a single logarithm that represents the logarithm of their product, expressed as \(\text{log}_{b}(m) + \text{log}_{b}(n) = \text{log}_{b}(mn)\).

Similarly, the quotient rule allows us to combine the difference of two logarithms into a single logarithm that represents the quotient of their contents, given as \(\text{log}_{b}(m) - \text{log}_{b}(n) = \text{log}_{b}(\frac{m}{n})\). When using these properties, it's vital to ensure all terms share the same base and to consider the domain of logarithmic functions, as logarithms are only defined for positive real numbers.

Exponential Form

Converting between logarithmic and exponential forms is a key step in solving logarithmic equations. The equation \(\text{log}_{b}(a) = c\) is equivalent to the exponential form \(b^c = a\). This conversion is based on the fundamental definition of logarithms.

In the context of solving the given exercise, transforming the logarithmic equation to its exponential counterpart simplifies the equation to a form that's easier to solve algebraically. This step is essential because it removes the logarithm and provides us with a standard algebraic equation. Remember, the base of the logarithm becomes the base of the exponent, the logarithmic term is equivalent to the exponent, and the other side of the equation becomes the result in the exponential form.

Factoring Quadratic Equations

When faced with a quadratic equation, factoring is a powerful tool used to find the solutions. A quadratic equation is typically written as \(ax^2+bx+c=0\). The process of factoring involves rewriting the quadratic as a product of binomials, which can then be solved using the zero-product property.

For the given equation \(x^2 - 7x - 8 = 0\), the step-by-step solution involves finding two numbers that multiply to give -8 (the constant term) and add to give -7 (the coefficient of \(x\)). In the exercise, the factored form is \((x - 8)(x + 1) = 0\), which leads to two potential values of \(x\) when each factor is set to zero. However, not all solutions obtained will necessarily satisfy the original equation's domain, and thus some solutions may need to be discarded, as shown in the final steps of the solution.