Problem 71 - 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. $$ \log _{2}(x+2)-\log _{2}(x-5)=3 $$

Step-by-Step Solution

Answer

The exact solution for the given logarithmic equation is \( x = 6 \).

Step 1: Simplify the equation using the properties of logarithms

Combine the two logarithmic terms on the left side of the equation using the quotient rule of logarithms which states that \(\log_bM - \log_bN = \log_b\frac{M}{N}\). Therefore, \(\log _{2}(x+2)-\log _{2}(x-5)=3\) can be simplified to \[\log _{2}\left(\frac{x+2}{x-5}\right)=3\]

Step 2: Convert the logarithmic equation to an exponential equation

Knowledge of the relationship between the logarithms and exponents is critical here. The equivalent exponential equation of \(\log _{b}(a)=c\) is \(b^{c}=a\). Using this relation, \(\log _{2}\left(\frac{x+2}{x-5}\right)=3\) becomes \[2^{3}=\frac{x+2}{x-5}\]

Step 3: Isolate the variable \(x\)

First simplify \(2^3\) which yields \(8\), thus this becomes \[8=\frac{x+2}{x-5}\]Then multiply both sides by \(x-5\) to get\[8(x-5)=x+2\]Simplify this to find the value of \(x\) by first expanding the left side and then gathering terms. This yields \[8x-40=x+27x=42\]Solve for \(x\) which gives \(x = 6\)

Step 4: Verify the solution

Before concluding, it is important check if the solution found is in the domain of the original logarithm expressions \( \log _{2}(x+2) \) and \( \log _{2}(x-5) \). For \( \log _{2}(x+2)\), the value under the logarithm (x+2) must be greater than \(0\).Hence, \(x+2 > 0\) which gives \(x>-2\)Thus, the solution \(x=6\) is valid for \( \log _{2}(x+2)\).For \( \log _{2}(x-5)\), the value under the logarithm (x-5) must be greater than \(0\).Hence, \(x-5 > 0\) which gives \(x>5\)Thus, the solution \(x=6\) is valid for \( \log _{2}(x-5)\).Since \(x = 6\) is in the domain of both original logarithm expressions, we accept this solution

Key Concepts

Properties of Logarithms

Logarithms have several important properties that simplify solving equations. One critical property is the quotient rule. This rule states that the difference between two logarithms with the same base can be expressed as a single logarithm of a fraction. More formally, \( \log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right) \). This property is particularly useful when solving equations like our example.

• Combining logarithms is often the first step to simplifying the equation.
• Using the quotient rule allows us to condense expressions and make the problem more manageable.

In the exercise, by applying this property, the problem reduces from two separate logarithmic expressions to one simplified form, making further steps easier.

Exponential Equations

Once logarithms are condensed, the next logical step is to convert the equation from logarithmic form to exponential form. The relationship here is that if \( \log_b(a) = c \), then \( b^c = a \). This conversion is fundamental because it switches the equation to a form that allows us to solve for the variable directly.For our example, \( \log_2\left(\frac{x+2}{x-5}\right) = 3 \) converts to \( 2^3 = \frac{x+2}{x-5} \). Solving exponential equations usually involves simple algebraic manipulation.

• First, evaluate the expression based on the base and power, such as \( 2^3 = 8 \).
• Next, isolate the variable by eliminating fractions or other terms through multiplication.

This methodology effectively takes advantage of the known equivalences between exponential and logarithmic expressions, simplifying the computations.

Domain of Logarithmic Functions

A crucial aspect when working with logarithms is ensuring that the values used are within the function's domain. Logarithms are only defined for positive numbers, meaning the expression inside a logarithm must be greater than zero.For both \( \log_2(x+2) \) and \( \log_2(x-5) \), we need to calculate conditions:

• \( x+2 > 0 \) leads to \( x > -2 \)
• \( x-5 > 0 \) leads to \( x > 5 \)

Thus, \( x \) must be greater than 5 for both conditions to be satisfied, ensuring the solution is valid.Checking the domain prevents us from accepting a solution that may cause mathematical errors or undefined expressions. Hence, validating the domain forms an essential part of solving logarithmic equations.

Solving Equations

Solving logarithmic equations often involves several key steps: applying properties of logarithms, converting to exponential form, simplifying, and checking the domain. Here's a summary of the process:

• Use properties such as the quotient rule to combine log terms.
• Convert the condensed logarithmic expression to an exponential equation.
• Solve for the variable through algebraic operations like distributing and collecting terms.
• Verify the solution by ensuring all expressions are within the domain of the logarithm.

This systematic approach not only finds a solution but ensures it's a valid one. Using these steps can empower students to tackle various logarithmic equations effectively and confidently.