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

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x = 6\).

Step 1: Simplify the equation

Using the properties of logarithms, we know that \(\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})\). So, we can rewrite the equation as \(\log _{2}(\frac{x+2}{x-5})=3\).

Step 2: Convert Logarithmic Equation to Exponential Equation

The core functionality of a logarithm \( \log_b(a) = c \) is the same as \( b^c = a \). Thus, the equation \(\log_2(\frac{x+2}{x-5} = 3) \) converts to \( 2^3 = \frac{x+2}{x-5}\). Simplifying this further gives \( 8 = \frac{x+2}{x-5}\).

Step 3: Solve for x

To find the value of \(x\), multiply both sides of the equation by \(x-5\) to get rid of the denominator. The result is \(8*(x-5) = x + 2\). Expanding and equating, we get \(8x - 40 = x + 2\). Rearranging this gives us \(7x = 42\). Solving for \(x\) ultimately yields \(x = 6\).

Step 4: Verify the Solution

In a logarithmic function, the domain is \(x > 0\), thus \(x+2 > 0\) and \(x-5 > 0\). In the provided solution, when \(x = 6\), it results in \(6+2 > 0\) and \(6-5 > 0\), both of which are true. So, \(x = 6\) is valid and within the domain of the original logarithmic expressions.

Key Concepts

Properties of Logarithms

When dealing with logarithmic equations, understanding the properties of logarithms is essential. These properties simplify complex logarithmic expressions, making them easier to solve. One key property is the difference of logs: \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \). This property allows us to combine two logarithmic expressions into one, as long as they have the same base.

There are other important properties to consider:

• Product Rule: \( \log_b(ab) = \log_b(a) + \log_b(b) \)
• Power Rule: \( \log_b(a^n) = n \cdot \log_b(a) \)
• Change of Base Formula: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \)

These rules simplify equations and help determine the steps to convert complex expressions into simpler forms. In our exercise, the crucial property used was the difference of logs, reducing \( \log_2(x+2) - \log_2(x-5) \) to a single log expression \( \log_2\left(\frac{x+2}{x-5}\right) \).

Exponential Equations

Exponential equations are closely related to logarithmic ones through the basic relationship \( \log_b(a) = c \) which is equivalent to \( b^c = a \). This relationship enables us to switch between exponential and logarithmic forms during problem-solving.

For example, after simplifying the logarithmic equation in our exercise to \( \log_2\left(\frac{x+2}{x-5}\right) = 3 \), we switched to its equivalent exponential form: \( 2^3 = \frac{x+2}{x-5} \). Performing this conversion helps unlock another method to solve the equation.

Exponential equations can sometimes be more straightforward to solve by isolating the variable. Converting a logarithmic equation into exponential form can make it easier to handle the arithmetic, leading us directly to the solution for \( x \).

Solving for x

Solving for \( x \) in any equation involves isolating \( x \) on one side. Once our logarithmic expression was rewritten as an exponential equation \( 8 = \frac{x+2}{x-5} \), the task was to simplify and solve for \( x \).

Here is how we approached it:

• First, multiply both sides by \( x-5 \) to eliminate the fraction: \( 8(x-5) = x+2 \).
• Expand the equation: \( 8x - 40 = x + 2 \).
• Rearrange terms to get \( 7x = 42 \).
• Finally, solve for \( x \) by dividing both sides by 7, resulting in \( x = 6 \).

These steps were crucial for isolating \( x \) and finding the solution. Once the arithmetic is clearly laid out, solving these equations becomes systematic.

Domain of Logarithmic Functions

A key aspect of solving logarithmic equations is understanding their domain. The domain is the set of all possible values of \( x \) for which the logarithmic expression is defined. For logarithms, the argument inside must always be greater than zero.

In our exercise, let's identify the domains of each logarithmic part:

• For \( \log_2(x+2) \), \( x+2 > 0 \) implies \( x > -2 \).
• For \( \log_2(x-5) \), \( x-5 > 0 \) implies \( x > 5 \).

Combining these conditions, we find the domain for the given logarithmic function is \( x > 5 \).

This means that any solution derived should respect these restrictions. In our solution, where \( x = 6 \), substitution into the expressions confirms they satisfy \( 6+2 > 0 \) and \( 6-5 > 0 \). Thus, \( x = 6 \) is validated as being within the function's domain. Understanding the domain helps ensure that we only choose valid solutions that won't lead to mathematical errors.