6.3. Example 3

🔹 Example 3: Solve: log𝑥 + log(𝑥 − 3) = 1

In this problem, we have two logarithmic expressions combined: log𝑥 and log(𝑥 - 3). They are set equal to 1. To solve this equation, we will use the Laws of Logarithms, specifically the Product Rule, which states that the sum of two logarithms is equal to the logarithm of the product of their arguments.

Think of it like combining two ingredients in a recipe. If you know how much of each ingredient you need when kept separately (log𝑥 for ingredient X and log(𝑥 - 3) for ingredient Y), you can combine them into one final product equation (the total amount needed).

Use product rule: log[𝑥(𝑥−3)] = 1

By applying the Product Rule, we can rewrite the equation log𝑥 + log(𝑥 - 3) as log[𝑥(𝑥 - 3)] = 1. This transformation makes it easier to solve for x since we have a single logarithmic expression rather than two separate ones.

Imagine you are mixing two paints together to create one color. Instead of keeping them separate (log𝑥 and log(𝑥 - 3)), you now have one mixed color (log[𝑥(𝑥 - 3)]). This makes it easier to work with and find the final outcome.

→ 𝑥(𝑥−3) = 10^1 = 10

After we have log[𝑥(𝑥 - 3)] = 1, we convert this logarithmic equation into exponential form. The equivalent exponential form indicates that the product 𝑥(𝑥 - 3) is equal to 10 (since 10 raised to the power of 1 is 10). This step transitions us from logarithmic language back to exponential language, allowing us to solve for x more directly.

If log is like asking the question 'how much of something do I have?', converting to exponential form is like answering that question. For example, if you have a combined paint worth of 10 units, you can directly look for the amounts of the two types of paint (x and x - 3) that give you that total.

→ 𝑥^2 - 3𝑥 - 10 = 0

We are now tasked with forming a quadratic equation. From 𝑥(𝑥 - 3) = 10, we expand it to create the equation 𝑥^2 - 3𝑥 - 10 = 0. Creating this quadratic equation allows us to use the quadratic formula or factoring to find the values of x that satisfy this equation.

Consider that you are setting up a scenario in an archery game to hit a specific target position (the number 10) with two arrows (x and x - 3). The combination of your arrows’ total ranges, represented by the quadratic, will help you evaluate the valid positions to hit that target.

→ (𝑥−5)(𝑥+2) = 0 → 𝑥 = 5 or 𝑥 = −2

Factoring the quadratic gives us two possible solutions: 𝑥 = 5 or 𝑥 = −2. This means that our equation has two critical points, both of which we need to analyze to check if they are valid solutions in the context of the original logarithmic expressions.

Imagine you are looking for two different paths to reach a target in a game. Both paths (their lengths represented by potential x values) seem valid initially, but you need to determine if both can lead you successfully towards your ultimate goal (the logarithmic context).

But 𝑥 = −2 is not valid (log of a negative number is undefined), Final answer: 𝑥 = 5

Since logarithms are undefined for negative numbers, we discard the solution x = -2. The only valid solution we keep is x = 5. This important step reveals how mathematical definitions constrain possible solutions, ensuring we only consider those that make sense in the logarithmic context.

Think about your possible paths again. If one path leads to an untraversable zone (like negative experience), you’d have to abandon that option. In contrast, a path leading to the target score (x = 5) remains viable and open for you.

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