Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the laws of logarithms to solve the equation. $$\log _{2} x-\log _{2}(x-2)=3$$

Step-by-Step Solution

Answer

The short answer to the given logarithmic equation is to combine the logs using the law of logarithm, remove the logarithm using the exponential rule, and solve for x. The solution is \(x = \frac{16}{7}\).

Step 1: Combine Logs using the Laws of Logarithm

Subtracting two logs of the same base is equivalent to taking the logarithm of the quotient of their arguments (the numbers inside the logarithm) using the law of logarithm: $$\log_{a}(x) - \log_{a}(y) = \log_{a}\left(\frac{x}{y}\right)$$ Applying this to the given equation: $$\log_{2}x - \log_{2}(x-2) = \log_{2}\left(\frac{x}{x-2}\right) = 3$$

Step 2: Remove the Logarithm using Exponential Rule

To remove the logarithm from both sides, we will raise both sides to the base (2) using the property: $$\log_{a}(x) = b \Rightarrow a^b = x$$ Using this property, we get: $$2^3 = \frac{x}{x-2}$$ Simplifying further, we get: $$8 = \frac{x}{x-2}$$

Step 3: Solve for x

To solve for x, first multiply both sides by the denominator (x-2): $$8(x-2) = x$$ Expanding and rearranging, we have: $$8x - 16 = x$$ Subtract x from both sides and add 16 to both sides to get: $$7x = 16$$ Now, divide both sides by 7 to solve for x: $$x = \frac{16}{7}$$ The solution to the given logarithmic equation is \(x = \frac{16}{7}\).

Key Concepts

Laws of Logarithms

Logarithms are a powerful mathematical tool that can help simplify complex equations, making them easier to solve. The laws of logarithms are rules that reflect the relationships between logarithms with the same base. These laws are crucial because they allow us to manipulate logarithmic expressions effectively.

Some of the key laws of logarithms include:

• Product Rule: The logarithm of a product is the sum of the logarithms of the factors. Mathematically, this is written as \( \log_{a}(xy) = \log_{a}(x) + \log_{a}(y) \).
• Quotient Rule: The logarithm of a quotient is the difference of the logarithms. This is shown by \( \log_{a}(\frac{x}{y}) = \log_{a}(x) - \log_{a}(y) \).
• Power Rule: The logarithm of a power is the exponent times the logarithm of the base number, expressed as \( \log_{a}(x^b) = b \cdot \log_{a}(x) \).

In the given exercise, we utilized the quotient rule. By applying this rule, we simplified \( \log_{2} x - \log_{2}(x-2) \) into \( \log_{2}(\frac{x}{x-2}) \). This step is critical in transforming the equation into a form that is solvable using exponential equations.

Exponential Equations

Exponential equations are equations where variables appear in the exponent. They often involve terms with bases and exponents and can be tricky to solve without the right approach.

The key property of logarithms that helps solve exponential equations is that the logarithm of a number is the exponent to which the base, a, must be raised to produce that number. This relationship is expressed as:

• \( \log_{a}(x) = b \implies a^b = x \)

For the given exercise, when we had \( \log_{2}(\frac{x}{x-2}) = 3 \), this allowed us to convert the logarithmic form into an exponential form. Using the property above, it became \( 2^3 = \frac{x}{x-2} \). Converting to exponential form is a handy trick that makes solving for variables in logarithmic equations manageable.

Solving Equations

Once we have used the laws of logarithms and converted the expression to an exponential equation, the next step is to solve the equation. Solving equations often requires algebraic manipulation to isolate the variable.

In the original problem, our goal was to find the value of \( x \). After converting to the exponential form \( 8 = \frac{x}{x-2} \), we had an equation with a simple proportion. Here's how we approached it:

• We multiplied both sides of the equation by the denominator \( (x-2) \) to eliminate the fraction: \( 8(x-2) = x \).
• Next, we expanded and simplified: \( 8x - 16 = x \).
• We then rearranged the equation to isolate \( x \) by subtracting \( x \) from both sides and adding 16: \( 7x = 16 \).
• Finally, we divided both sides by 7 to solve for \( x \): \( x = \frac{16}{7} \).

This sequence of steps, starting from applying logarithmic laws to isolating the variable, illustrates how to methodically tackle logarithmic and exponential equations, leading to a clear solution.