Master Logarithms: Solving Complex Equations

Hey everyone, let's dive deep into the awesome world of logarithms today, guys! We're going to tackle some seriously cool problems that will really flex your math muscles. Logarithms might seem a bit intimidating at first, but once you get the hang of the basic rules, they become super manageable and even fun to work with. Think of them as the inverse operation to exponentiation – they help us figure out what power we need to raise a base to in order to get a certain number. We'll be going through a bunch of examples, starting with combining logarithms using addition and subtraction, then moving on to problems involving coefficients. Get ready to boost your understanding and confidence with these logarithmic expressions!

Combining Logarithms with Addition and Subtraction

Alright, first up, we're going to explore how to combine logarithmic expressions when we're adding or subtracting them. This is where some of the most fundamental logarithm properties come into play, and understanding these is key to solving more complex problems down the line. Remember these golden rules, guys: when you add two logarithms with the same base, you can multiply their arguments (the numbers inside the log), and when you subtract them, you can divide their arguments. It's like magic, but it's just pure math!

Example c) log₃16 + log₃9

Let's kick things off with a classic addition problem: log₃16 + log₃9. See how both logarithms have the same base, which is 3? That's our cue to use the product rule for logarithms. This rule states that log<0xE2><0x82><0x99>(a) + log<0xE2><0x82><0x99>(b) = log<0xE2><0x82><0x99>(ab). So, in our case, we can combine these two logs into a single logarithm by multiplying the numbers inside: log₃(16 * 9). Now, we just do the multiplication: 16 * 9 = 144. So, the expression simplifies to log₃144. This is a great example of how we can condense multiple logarithmic terms into one, making it much simpler to work with.

Example d) log₆24 + log₆9

Following the same logic, let's tackle log₆24 + log₆9. Again, we have the same base, 6, for both logarithms. This means we can apply the product rule: log₆(24 * 9). Multiplying 24 by 9 gives us 216. So, the expression simplifies to log₆216. Now, here's a little bonus step we can take: can you think of what power you need to raise 6 to in order to get 216? That's right, it's 3 (since 6 * 6 * 6 = 216). So, log₆216 = 3. See how simplifying can lead to a nice, clean numerical answer sometimes? It's always worth checking if the resulting logarithm can be evaluated to a simple number.

Example e) log₃12 - log₃4

Now, let's switch gears to subtraction. We've got log₃12 - log₃4. When we subtract logarithms with the same base, we use the quotient rule, which says log<0xE2><0x82><0x99>(a) - log<0xE2><0x82><0x99>(b) = log<0xE2><0x82><0x99>(a/b). So, we can rewrite our expression as log₃(12 / 4). Performing the division, 12 / 4 = 3. This gives us log₃3. And what power do you raise 3 to get 3? Easy peasy, it's 1! So, log₃3 = 1. This example shows that subtraction can often simplify things down to a very basic answer.

Example f) log₄18 - log₄9

Let's do another subtraction one: log₄18 - log₄9. Same base (4), so we use the quotient rule again. This becomes log₄(18 / 9). Dividing 18 by 9 gives us 2. So, the expression simplifies to log₄2. This one might not simplify to a whole number immediately, but it's a perfectly valid simplified form. If you needed to approximate it, you'd use a calculator, but for exact answers, log₄2 is great. Think about it: what power of 4 gives you 2? It's 1/2, because 4^(1/2) is the square root of 4, which is 2. So, log₄2 = 1/2.

Example g) log₉4 + log₉0.25

Here's a fun one: log₉4 + log₉0.25. We've got addition and the same base (9). So, we multiply the arguments: log₉(4 * 0.25). Now, 0.25 is the same as 1/4. So, we're calculating 4 * (1/4). Anything multiplied by its reciprocal equals 1! So, 4 * 0.25 = 1. Our expression becomes log₉1. And here's another super important logarithm rule: the logarithm of 1 to any base (except 1) is always 0. This is because any non-zero number raised to the power of 0 equals 1. So, log₉1 = 0. Remember this one, guys – it pops up a lot!

Working with Coefficients and Logarithm Properties

Now that we've got the hang of adding and subtracting, let's level up by introducing coefficients. Coefficients in front of logarithms, like the '2' in 2 log₁₀2, represent powers. The power rule for logarithms states that n log<0xE2><0x82><0x99>(a) = log<0xE2><0x82><0x99>(aⁿ). This rule is super powerful because it allows us to move a coefficient inside the logarithm as an exponent. This is often the first step in simplifying expressions that have coefficients, as it allows us to then apply the product or quotient rules if needed. We'll also see how understanding basic logarithm values, like log<0xE2><0x82><0x99>(b) = 1 when b=a, and log<0xE2><0x82><0x99>(1) = 0, remains crucial for getting to the final, simplest answer. Get ready, because these problems combine multiple rules!

Example h) 2 log₁₀2 + log₁₀25

Let's tackle 2 log₁₀2 + log₁₀25. First, look at the term 2 log₁₀2. The '2' is a coefficient. Using the power rule, we can move this '2' inside the logarithm as an exponent: log₁₀(2²). That simplifies to log₁₀4. So, our original expression is now log₁₀4 + log₁₀25. We're back to adding logarithms with the same base (10)! So, we use the product rule: log₁₀(4 * 25). Calculating 4 * 25 gives us 100. So, we have log₁₀100. Now, what power do you raise 10 to in order to get 100? That's right, it's 2 (since 10 * 10 = 100). Therefore, log₁₀100 = 2. Another great simplification achieved!

Example i) (1/3) log₃3 - log₃18 + log₃9

This one looks a bit more complex: (1/3) log₃3 - log₃18 + log₃9. Let's break it down step-by-step, guys. First, deal with the coefficient (1/3) log₃3. Using the power rule, this becomes log₃(3^(1/3)). Now, we have log₃(3^(1/3)) - log₃18 + log₃9. All have the same base (3), so we can combine them. Remember, subtraction means division, and addition means multiplication. Let's group the positive terms first: log₃(3^(1/3)) + log₃9. This combines to log₃(3^(1/3) * 9). Now, we subtract log₃18, which means we divide by 18: log₃( (3^(1/3) * 9) / 18 ). Let's simplify the argument: (3^(1/3) * 9) / 18. We know 9/18 simplifies to 1/2. So, the argument is 3^(1/3) * (1/2). This gives us log₃( (1/2) * 3^(1/3) ). While this is a mathematically correct simplification, it doesn't simplify to a nice integer or simple fraction. Let's re-examine the original expression carefully. Often, problems like these are designed to simplify nicely. Let's re-evaluate the terms. We have (1/3)log₃3. Since log₃3 = 1, this term is simply (1/3) * 1 = 1/3. So the expression is 1/3 - log₃18 + log₃9. Now, let's combine the last two terms: -log₃18 + log₃9. This is the same as log₃9 - log₃18, which by the quotient rule is log₃(9/18) = log₃(1/2). So our expression is 1/3 + log₃(1/2). This is a valid simplified form. However, if the intention was for a cleaner numerical answer, let's consider if there might be a typo or a different interpretation. Assuming the original intent was to simplify as much as possible using logarithm rules, 1/3 + log₃(1/2) is the result. If we wanted to combine all terms into a single log, we'd have to express 1/3 as a log, which is log₃(3^(1/3)). Then we'd have log₃(3^(1/3)) + log₃(1/2) = log₃( (3^(1/3))/2 ). This highlights that sometimes the