6. Solving Logarithmic Equations

🔹 Example 1: Solve: log 𝑥 = 5

→ Convert to exponential form: 𝑥 = 25 = 32

In this example, we start with the logarithmic equation: log 𝑥 = 5. To solve for 𝑥, we need to rewrite the equation in its exponential form. The notation 'log 𝑥 = 5' tells us that 𝑥 is the number we get when we raise the base (which is 10, if not specified) to the power of 5. This means:

• If log 𝑥 = 5, then 𝑥 = 10^5.
• So, 𝑥 = 100000, which can also be simplified to express that 10 raised to the power of 5 equals x.
• For clarification, converting logarithmic equations back to their exponential forms is key to finding the unknowns in logarithmic statements.

Think of logarithmic functions like a special kind of detective story where you are looking for missing information. Just like detectives have their clues (the logs), they have to convert the information into a different form to figure out what happened. In solving our equation 𝑥 = 10^5, it's like uncovering a clue that reveals how 'large' our answer is. The larger the exponent, like 5 in this case, the bigger the mystery unfolds because we realize we're dealing with a very large number!

🔹 Example 2: Solve: log (𝑥+1) = 2

→ 𝑥+1 = 32 = 9 → 𝑥 = 8

For this example, we start with log (𝑥 + 1) = 2. To solve for 𝑥, we convert the logarithmic statement to exponential form. It tells us that:

• 𝑥 + 1 = 10^2,
• Which simplifies to 𝑥 + 1 = 100.
• Now, we can solve for 𝑥 by subtracting 1 from both sides:
• 𝑥 = 100 - 1,
• Thus, 𝑥 = 99. This is an application of using logarithms to find values that are represented in a different way.

Imagine you are accumulating points in a video game, and needing to reach a certain level. The game level requirement states that you need to collect 10^2 or 100 experience points to level up. This problem resembles how we worked through our equation by determining how much more you needed to achieve a level by simplifying what you had (the +1), ultimately determining your exact score needed to achieve that great transformation into the next level!

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

Use product rule: log[𝑥(𝑥−3)] = 1 → 𝑥(𝑥−3) = 10^1 = 10 → 𝑥^2 − 3𝑥 − 10 = 0 → Solve quadratic: (𝑥−5)(𝑥+2) = 0 → 𝑥 = 5 or 𝑥 = −2 But 𝑥 = −2 is not valid (log of negative number is undefined), Final answer: 𝑥 = 5

In this example, we have two logarithmic terms combined: log𝑥 + log(𝑥 − 3) = 1. To solve this, we can utilize the product rule of logarithms, which states that the sum of logs is the log of the product of those numbers. By applying this rule, we write:

• log[𝑥(𝑥−3)] = 1.

Next, we convert this back to exponential form:

• 𝑥(𝑥−3) = 10^1, which simplifies to 𝑥(𝑥−3) = 10.
• This leads us to the quadratic equation: 𝑥^2 − 3𝑥 − 10 = 0.
• Solving the quadratic gives us two potential solutions: 𝑥 = 5 or 𝑥 = −2. However, we must discard 𝑥 = −2 because logarithms cannot take a negative input. Thus, the valid solution is 𝑥 = 5.

Consider an instance where you're mixing two ingredients to achieve a specific flavor (like the terms log𝑥 and log(𝑥−3)). The total desired outcome (1) can be thought of as the result of combining those ingredients. If we think of each part (the logs) representing a recipe ratio, applying the product rule helps ensure the correct balance is achieved in the cooking process. Thus, our final solution (5) represents a perfectly balanced flavor, while any negative mixing (like -2) would spoil the dish and is not acceptable!

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