Problem 2 - Algebra and Trigonometry

Question

Let's solve the logarithmic equation \(\log 3+\log (x-2)=\log x\). (a) First, we combine the logarithms to get the equivalent equation __________. (b) Next, we write each side in exponential form to get the equivalent equation __________. (c) Now we find \(x=\) __________.

Step-by-Step Solution

Answer

Combine logs: \( \log(3(x-2)) = \log x \); Solve: \(x = 3\).

Step 1: Combine Logarithms

First, we use the property of logarithms: \( \log a + \log b = \log(ab) \). Applying this property to the left side of the equation, we have \( \log 3 + \log (x-2) = \log(3(x-2)) \). Thus, the equation becomes \( \log(3(x-2)) = \log x \).

Step 2: Remove Logarithms via Exponential Form

Next, we can remove the logarithms on both sides by equating the arguments since the logarithm is a one-to-one function. Therefore, we have: \( 3(x-2) = x \).

Step 3: Solve the Linear Equation

Now, solve the linear equation for \(x\) by distributing and re-arranging the terms. Start by expanding: \( 3x - 6 = x \). Move \(x\) to one side: \(3x - x = 6\). Simplify to get \(2x = 6\). Divide both sides by 2 to isolate \(x\), giving \(x = 3\).

Step 4: Check the Solution

Substitute \(x = 3\) back into the original equation to ensure it does not produce any undefined expressions, particularly ensuring the logarithm arguments are positive. Both sides become \(\log(3) + \log(1)\), confirming \(x = 3\) is valid and satisfies the original equation.

Key Concepts

Properties of Logarithms

Logarithms have several key properties that are very useful in solving equations. One of the most important is the product rule, which states:

• \( \log a + \log b = \log(ab) \)

This property allows us to combine two logarithms into one, which simplifies solving equations involving logarithms. In our problem, we apply this by changing \( \log 3 + \log(x-2) \) into \( \log(3(x-2)) \). By simplifying the left-hand side, we make it easier to equate the expressions within the logarithms.
Another property is the one-to-one nature of logarithms, which tells us that if \( \log a = \log b \), then \( a = b \). This allows us to drop the logs once they're combined, simplifying the equation further to a simple form that can be solved without dealing with logarithms.

Exponential Form

The exponential form is a way of rewriting logarithmic equations to further simplify solving them. Logarithms and exponents are inverse operations, which means they "undo" each other. A logarithmic expression like \( \log_b(a) = c \) can be rewritten in exponential form as \( b^c = a \). In essence, if you understand one, you can easily convert to the other.
In the context of our problem, after combining the logs using properties of logarithms, we have the equation \( \log(3(x-2)) = \log x \). This can then be interpreted in exponential form to simply equate what's inside the logs: \( 3(x-2) = x \). By converting to exponential form, we remove the logs entirely, thereby transitioning the problem into a linear equation.

Solving Linear Equations

Once we have removed the logarithms, we find ourselves with a linear equation \( 3(x-2) = x \). Solving linear equations involves simple algebraic manipulations to isolate the unknown variable. Here is how you can solve the equation step-by-step:

• Distribute the 3 across the terms inside the parentheses, giving us \( 3x - 6 = x \).
• Next, gather all terms involving \( x \) on one side of the equation. So, subtract \( x \) from both sides: \( 3x - x = 6 \).
• Simplify the expression to \( 2x = 6 \).
• Finally, divide both sides by 2 to isolate \( x \), which provides the solution: \( x = 3 \).

Always check your solution by substituting \( x = 3 \) back into the original equation to verify it satisfies all conditions, especially that the argument of any logarithm needs to be positive. In this case, the solution keeps all terms defined and confirms that \( x = 3 \) is correct.