Problem 76 - Algebra and Trigonometry

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

Step-by-Step Solution

Answer

The exact solutions and approximations depend on the solutions of the quadratic equation in step 3. Be sure to double-check these values by substituting them back into the original equation, \(\log _{2}(x-3)+\log _{2} x-\log _{2}(x+2)=2\), since not all values calculated will be valid solutions.

Step 1: Combine the logarithms

Using the logarithm properties \(\log_b m + \log_b n = \log_b (mn)\) and \(\log_b m - \log_b n = \log_b (m/n)\), to combine the logarithms into one: \(\log_2\((x-3)x/(x+2)) = 2\).

Step 2: Convert log to exponential

In order to solve for \(x\), we convert the given equation from logarithmic form to exponential form:\(2^2 = (x-3)x/(x+2)\).

Step 3: Simplify and solve

First, simplify the equation:4 = \(x^2 - 3x / (x+2)\). The next steps involve manipulating this equation to get a quadratic equation. Multiply both sides by \((x+2)\) to get rid of the denominator:\(4(x+2) = x^2 - 3x\). Then simplify this equation to a standard quadratic equation and solve for \(x\) using the quadratic formula. The resulting equation is: \(x^2 - 3x - 4(x+2) = 0\).

Step 4: Check the solutions

After solving the quadratic equation you would hopefully get two solutions for \(x\). For each value, substitute it back into the original logarithmic equation to check if it satisfies the equality and is in the domain of the original problem. Remember that, in the logarithmic expression \(\log_b(x)\), \(x\) must be positive. So any value that is not positive should be rejected.

Step 5: Decimal approximation

After getting the acceptable exact solutions of this equation, use a calculator to obtain a decimal approximation for each solution, correct to two decimal places.

Key Concepts

Logarithm Properties

Logarithms have several important properties that make them extremely useful, especially in solving equations. One key property is the product rule. This states that the logarithm of a product is equal to the sum of the logarithms: \[ \log_b(mn) = \log_b(m) + \log_b(n) \] This is useful when combining logarithms into a single expression.
Another important property is the quotient rule, which says that the logarithm of a quotient is the difference of the logarithms: \[ \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \] Understanding these rules will help you manipulate and simplify logarithmic equations.
In the given problem, these logarithm properties are used to combine and simplify the expression, making it easier to solve using other algebraic methods like converting to exponential form.

Exponential Form

Transitioning from a logarithmic equation to exponential form is a crucial step in solving logarithmic equations. The general idea is to "undo" the logarithm by expressing the log equation as an exponent. For any logarithmic equation of the form:\[ \log_b(a) = c \]It is equivalent to saying:\[ b^c = a \]
In the step-by-step solution, the equation was converted from \( \log_2((x-3)x/(x+2)) = 2 \) into its exponential form: \[ 2^2 = \frac{(x-3)x}{x+2} \] This transformation is vital as it reveals a more familiar equation format that we can solve using algebraic techniques.
The exponential form eliminates the logarithm and helps in handling equations involving powers and roots.

Quadratic Equation

Once the equation has been transformed into exponential form, simplifying it often leads to a quadratic equation, a fundamental element of algebra. A quadratic equation is generally presented in the standard form:\[ ax^2 + bx + c = 0 \] To solve the quadratic equation, you can use various methods such as factoring, completing the square, or applying the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the exercise, the equation was simplified to \( 4(x+2) = x^2 - 3x \). This required converting it into a standard quadratic form to solve for \(x\). Solving the quadratic equation provides potential solutions that must be checked against the conditions of the problem.
The quadratic formula is often used because it reliably finds solutions for any set of coefficients \(a\), \(b\), and \(c\).

Domain of Logarithmic Functions

One critical aspect of logarithmic functions is understanding their domain. The domain of a logarithmic function, like \( \log_b(x) \), is all positive real numbers \(x > 0\). This is because the logarithm of a non-positive number (zero or negative) is undefined.
When solving logarithmic equations, it's imperative to verify that the solutions fall within the acceptable domain.

• The values substituted back must keep the argument of every logarithm positive.
• Any result outside this domain should be discarded.

In the exercise, once potential solutions for \(x\) were found by solving the quadratic equation, these solutions needed to be checked back against the original expression: \[ \log _{2}(x-3)+\log _{2} x-\log _{2}(x+2)=2 \] This was to ensure every log term involves a positive argument, thereby confirming it is within the domain. Neglecting this check can lead to including invalid solutions.