Solving Logarithmic Expressions: A Step-by-Step Guide

October 9, 2025 by TextBrain Team 54 views

Hey guys! Ever get those tricky logarithmic expressions that seem like a jumbled mess? Don't worry, we've all been there! Today, we're going to break down one of those problems step-by-step, making it super easy to understand. We'll tackle this expression: ⁴log 9 ⋅ ³log 125 ⋅ ²⁵log 16. Let's dive in and make sense of it together!

Understanding Logarithms

Before we jump into solving this expression, let's quickly refresh our understanding of logarithms. Think of a logarithm as the inverse operation of exponentiation. In simple terms, if we have an expression like b^x = y, the logarithmic form of this expression is log_b(y) = x. Here:

b is the base of the logarithm.
y is the argument (the value we're taking the logarithm of).
x is the exponent to which we must raise the base b to get y.

For example, log₂8 = 3 because 2³ = 8. The logarithm answers the question: “To what power must we raise the base (2) to get the argument (8)?” In this case, the answer is 3.

The beauty of logarithms lies in their properties, which allow us to simplify complex expressions. Here are a couple of key properties we'll be using today:

Change of Base Formula: log_a(b) = log_c(b) / log_c(a) This property lets us change the base of a logarithm, which is super handy when dealing with different bases in the same expression.
Power Rule: log_a(bⁿ) = n ⋅ log_a(b) This rule allows us to bring exponents out of the logarithm, simplifying our calculations.

With these basics in mind, we're well-equipped to tackle the expression at hand. Remember, logarithms might seem intimidating at first, but with a bit of practice, they become much more manageable. So, let's move on and see how we can apply these concepts to our specific problem!

Breaking Down the Expression: ⁴log 9

Okay, let's kick things off by focusing on the first part of our expression: ⁴log 9. This might look a bit daunting at first, but don't sweat it! We're going to break it down into smaller, more manageable pieces.

The key here is to recognize that both 4 and 9 can be expressed as powers of a common base. In this case, both are powers of 2 and 3 respectively, but we can dig deeper. Notice that 4 is 2 squared (2²) and 9 is 3 squared (3²). This is a crucial observation because it allows us to use the properties of logarithms to simplify the expression. We can rewrite ⁴log 9 using these powers:

⁴log 9 = log₂² 3²

Now we can use the power rule of logarithms, which states that log_a(bⁿ) = n ⋅ log_a(b). Applying this rule, we get:

log₂² 3² = (2 ⋅ log₂ 3) / (2 ⋅ log₂ 2)

See what we did there? We essentially brought the exponents (the powers) out in front of the logarithms. This is a super helpful trick for simplifying expressions. Next, we can simplify further.

Since log₂ 2 is simply equal to 1 (because 2 raised to the power of 1 is 2), we can simplify the expression to:

(2 ⋅ log₂ 3) / (2 ⋅ log₂ 2) = (2 ⋅ log₂ 3) / (2 ⋅ 1) = log₂ 3

So, ⁴log 9 simplifies beautifully to log₂ 3. This is a huge step forward! By recognizing the common base and applying the power rule, we've transformed a seemingly complex term into something much simpler. This is the kind of approach we'll use throughout the problem.

Now, let's recap. We started with ⁴log 9, recognized that 4 and 9 are powers of smaller numbers, applied the power rule of logarithms, and simplified the expression to log₂ 3. This might seem like a lot of steps, but each step is logical and brings us closer to the solution. Next, we'll tackle the second part of the expression, ³log 125, using a similar approach. Stay tuned!

Simplifying the Next Term: ³log 125

Alright, let's move on to the second part of our expression: ³log 125. Just like before, we want to see if we can express the base and the argument as powers of a common number. In this case, the base is 3, and the argument is 125.

Think about 125 for a moment. Can we write it as a power of 3? Nope! But can we write it as a power of something else? Absolutely! 125 is 5 cubed (5³). This is a crucial insight because it allows us to simplify the logarithm using the power rule. So, we can rewrite ³log 125 as:

³log 125 = ³log 5³

Now, we can apply the power rule of logarithms, which, as we remember, states that log_a(bⁿ) = n ⋅ log_a(b). Applying this rule, we get:

³log 5³ = 3 ⋅ ³log 5

And that's it! We've simplified ³log 125 to 3 ⋅ ³log 5. Notice how we didn't try to force a base of 3 – instead, we looked for the most natural base for the argument (125), which was 5. This is a key strategy when dealing with logarithms: find the simplest representation for each part of the expression.

So, just to recap, we started with ³log 125, recognized that 125 is 5³, and applied the power rule to simplify the expression to 3 ⋅ ³log 5. We're making great progress! We've now simplified two out of the three terms in our original expression. Next up, we'll tackle the final term, ²⁵log 16, using the same principles. Keep the momentum going!

Tackling the Final Term: ²⁵log 16

Okay, guys, we're on the home stretch! Let's tackle the last piece of our puzzle: ²⁵log 16. This one might look a little trickier at first glance, but we'll break it down just like the others. Remember our strategy: try to express both the base and the argument as powers of a common number.

Let's start with the base, 25. We can express 25 as 5 squared (5²). Now, let's look at the argument, 16. We can express 16 as 2 to the power of 4 (2⁴). So, we can rewrite ²⁵log 16 as:

²⁵log 16 = log₅² 2⁴

Now we can use the power rule of logarithms. Remember, log_a(bⁿ) = n ⋅ log_a(b). Applying this rule, we get:

log₅² 2⁴ = (4 ⋅ log₅ 2) / (2 ⋅ log₅ 5)

Notice that we've applied the power rule to both the base and the argument. We brought the exponent of the argument (4) out in front, and we also considered the exponent of the base (2) when rewriting the logarithm. Now, let's simplify.

Since log₅ 5 is simply equal to 1 (because 5 raised to the power of 1 is 5), we can simplify the expression to:

(4 ⋅ log₅ 2) / (2 ⋅ log₅ 5) = (4 ⋅ log₅ 2) / (2 ⋅ 1) = 2 ⋅ log₅ 2

So, ²⁵log 16 simplifies to 2 ⋅ log₅ 2. Excellent! We've now simplified all three individual logarithmic terms in our original expression. We're in a fantastic position to put it all together and find the final answer.

Just to recap, we started with ²⁵log 16, expressed 25 as 5² and 16 as 2⁴, applied the power rule of logarithms, and simplified the expression to 2 ⋅ log₅ 2. We've conquered the individual pieces; now let's combine them and see the magic happen!

Putting It All Together: The Final Calculation

Alright, guys, this is where the magic happens! We've simplified each part of our original expression, and now it's time to put it all together and calculate the final result. Let's recap what we've found so far:

⁴log 9 = log₂ 3
³log 125 = 3 ⋅ ³log 5
²⁵log 16 = 2 ⋅ log₅ 2

Our original expression was ⁴log 9 ⋅ ³log 125 ⋅ ²⁵log 16. Now we can substitute our simplified expressions:

⁴log 9 ⋅ ³log 125 ⋅ ²⁵log 16 = (log₂ 3) ⋅ (3 ⋅ ³log 5) ⋅ (2 ⋅ log₅ 2

Now, let's rearrange the terms to make things a bit clearer:

(log₂ 3) ⋅ (3 ⋅ ³log 5) ⋅ (2 ⋅ log₅ 2) = 3 ⋅ 2 ⋅ (log₂ 3) ⋅ (³log 5) ⋅ (log₅ 2) = 6 ⋅ (log₂ 3) ⋅ (³log 5) ⋅ (log₅ 2)

Notice that we've just rearranged the constants (3 and 2) to the front. Now, we have a beautiful sequence of logarithms. To simplify this further, we'll use the change of base formula. This formula is a game-changer because it allows us to switch the base of a logarithm to something more convenient. The change of base formula states:

log_a(b) = log_c(b) / log_c(a)

We're going to apply this formula strategically to our expression. Let's focus on the last two logarithmic terms: (³log 5) ⋅ (log₅ 2). We can rewrite ³log 5 using the change of base formula with base 2:

³log 5 = log₂ 5 / log₂ 3

Now, let's substitute this back into our expression:

6 ⋅ (log₂ 3) ⋅ (log₂ 5 / log₂ 3) ⋅ (log₅ 2)

See anything cool happening? We have a log₂ 3 in the numerator and a log₂ 3 in the denominator! They cancel each other out:

6 ⋅ (log₂ 3) ⋅ (log₂ 5 / log₂ 3) ⋅ (log₅ 2) = 6 ⋅ (log₂ 5) ⋅ (log₅ 2)

Now, let's apply the change of base formula again, this time to log₅ 2, changing the base to 2:

log₅ 2 = log₂ 2 / log₂ 5

Substitute this back into our expression:

6 ⋅ (log₂ 5) ⋅ (log₂ 2 / log₂ 5)

Again, we see some cancellation magic! We have a log₂ 5 in the numerator and a log₂ 5 in the denominator. They cancel out, leaving us with:

6 ⋅ (log₂ 5) ⋅ (log₂ 2 / log₂ 5) = 6 ⋅ log₂ 2

And we know that log₂ 2 is simply 1 (because 2 raised to the power of 1 is 2). So our expression simplifies to:

6 ⋅ log₂ 2 = 6 ⋅ 1 = 6

Conclusion: The Grand Finale

And there you have it, guys! We started with the seemingly complex expression ⁴log 9 ⋅ ³log 125 ⋅ ²⁵log 16, and after a step-by-step journey through the properties of logarithms, we've arrived at the final answer: 6. How awesome is that?

Let's take a moment to appreciate the process we went through. We didn't just blindly try to plug numbers into a calculator. Instead, we:

Broke down the problem: We divided the expression into smaller, manageable parts.
Identified common bases: We looked for ways to express the bases and arguments of the logarithms as powers of common numbers.
Applied the power rule: We used the power rule to simplify the logarithms.
Used the change of base formula: We strategically changed the bases of the logarithms to create opportunities for cancellation.
Simplified and combined: We carefully simplified each term and then combined them to arrive at the final answer.

This approach is not just about getting the right answer; it's about understanding the why behind the math. By understanding the properties of logarithms and how to apply them, you can tackle all sorts of logarithmic expressions with confidence.

So, next time you encounter a tricky logarithmic problem, remember the steps we've covered today. Break it down, look for common bases, apply the power rule, use the change of base formula, and simplify. You've got this! And remember, math can be fun – especially when you're conquering complex problems like this one. Keep practicing, keep exploring, and keep those mathematical muscles strong! You guys are awesome, and I'm so glad we could tackle this together. Until next time, happy calculating!