Logarithm Calculations: Step-by-Step Solutions

Hey guys! Today, we're diving into the exciting world of logarithms! We'll tackle some problems that might look a little intimidating at first, but trust me, with a little understanding, they're totally manageable. We're going to break down the calculations step-by-step, so you can follow along and boost your math skills. Let's get started!

Understanding Logarithms

Before we jump into solving the problems, let's quickly refresh what logarithms are all about. Think of a logarithm as the inverse operation of exponentiation. Basically, if we have an equation like bˣ = y, the logarithm (with base b) of y is x. We write this as log_b(y) = x. So, the logarithm answers the question: "To what power must we raise the base (b) to get y?"

Key Properties of Logarithms:

• Product Rule: log_b(mn) = log_b(m) + log_b(n)
• Quotient Rule: log_b(m/n) = log_b(m) - log_b(n)
• Power Rule: log_b(m^p) = p * log_b(m)
• Change of Base Formula: log_a(b) = log_c(b) / log_c(a) (This one is super handy!)

These properties are our secret weapons for simplifying and solving logarithmic expressions. Keep them in mind as we work through the examples.

Problem 1: Calculating ³log 4 * ⁵log 27

Okay, let's dive into our first problem: ³log 4 * ⁵log 27. This looks a bit complex, but we can break it down using the properties we just discussed. Our main goal here is to simplify each logarithmic term and see if we can find any common ground.

First, let's think about rewriting the numbers inside the logarithms as powers of prime numbers. This is a very useful trick in logarithm problems. We know that 4 can be written as 2², and 27 can be written as 3³. So, let's rewrite our expression:

³log 4 * ⁵log 27 = ³log (2²) * ⁵log (3³)

Now we can use the power rule of logarithms, which states that log_b(m^p) = p * log_b(m). Applying this rule to both terms, we get:

= 2 * ³log 2 * 3 * ⁵log 3

= 6 * ³log 2 * ⁵log 3

At this point, it might not be immediately obvious how to proceed. This is where the change of base formula comes to the rescue! Remember, the change of base formula allows us to convert a logarithm from one base to another. We can choose a convenient base, like 10 or the natural base e, to make things simpler. Let's choose base 10 for this example.

Using the change of base formula, we can rewrite ³log 2 and ⁵log 3 as:

³log 2 = log 2 / log 3

⁵log 3 = log 3 / log 5

Now, substitute these back into our expression:

6 * ³log 2 * ⁵log 3 = 6 * (log 2 / log 3) * (log 3 / log 5)

Notice anything cool? We have a 'log 3' in the numerator and the denominator, so they cancel each other out! This leaves us with:

= 6 * (log 2 / log 5)

We can leave the answer in this form, or we can use a calculator to get a decimal approximation. Using a calculator, we find that log 2 ≈ 0.3010 and log 5 ≈ 0.6990. So,

6 * (log 2 / log 5) ≈ 6 * (0.3010 / 0.6990) ≈ 6 * 0.4306 ≈ 2.5836

So, the final answer for the first part is approximately 2.5836. Remember, it's important to show your steps clearly, so you can easily track your work and spot any potential errors. Understanding the properties of logarithms is key to solving these types of problems.

Problem 2: Calculating ⁵log 4 * ²log 16

Let's move on to the second part of the problem: ⁵log 4 * ²log 16. Again, our strategy is to simplify each term using the properties of logarithms. We'll start by expressing the numbers inside the logarithms as powers of prime numbers.

We know that 4 is 2², and 16 is 2⁴. So, let's rewrite the expression:

⁵log 4 * ²log 16 = ⁵log (2²) * ²log (2⁴)

Now we use the power rule of logarithms, which says log_b(m^p) = p * log_b(m). Applying this to both terms, we get:

= 2 * ⁵log 2 * 4 * ²log 2

= 8 * ⁵log 2 * ²log 2

Now, this looks promising! We have two logarithmic terms, but they have different bases. To make things easier, we'll use the change of base formula again. This time, let's convert both logarithms to base 10.

⁵log 2 = log 2 / log 5

²log 2 = log 2 / log 2 = 1 (Remember: log_b(b) = 1)

Substitute these back into the expression:

8 * ⁵log 2 * ²log 2 = 8 * (log 2 / log 5) * 1

= 8 * (log 2 / log 5)

Just like before, we can use a calculator to find the approximate values of log 2 and log 5. We already know that log 2 ≈ 0.3010 and log 5 ≈ 0.6990. So,

8 * (log 2 / log 5) ≈ 8 * (0.3010 / 0.6990) ≈ 8 * 0.4306 ≈ 3.4448

So, the answer for the second part is approximately 3.4448. Remember, the key to these problems is to break them down into smaller, manageable steps. Don't be afraid to use the properties of logarithms – they are your best friends!

Problem 3: Calculating ²⁵log √10

Let's tackle the final part of the problem: ²⁵log √10. This one involves a square root, but don't worry, we've got this! Remember that a square root can be expressed as a fractional exponent. So, √10 is the same as 10^(1/2).

Let's rewrite the expression using this fact:

²⁵log √10 = ²⁵log (10^(1/2))

Now, we can use the power rule of logarithms again. This gives us:

= (1/2) * ²⁵log 10

Okay, now we need to think about the bases. We have a base of 25, and we want to relate it to 10 somehow. Notice that 25 is 5², and 10 is 2 * 5. This might give us a clue about how to proceed. Let's rewrite 25 as 5²:

= (1/2) * ⁵²log 10

Now, let's use a lesser-known but very useful property of logarithms: log_(a^b)(c) = (1/b) * log_a(c). This property helps us when the base of the logarithm has an exponent. Applying this property, we get:

= (1/2) * (1/2) * ⁵log 10

= (1/4) * ⁵log 10

Now, we need to express ⁵log 10 in terms of simpler logarithms. We can use the fact that 10 = 2 * 5 and the product rule of logarithms:

⁵log 10 = ⁵log (2 * 5) = ⁵log 2 + ⁵log 5

Remember that log_b(b) = 1, so ⁵log 5 = 1. This simplifies our expression to:

⁵log 10 = ⁵log 2 + 1

Substitute this back into our main expression:

(1/4) * ⁵log 10 = (1/4) * (⁵log 2 + 1)

Now, we can use the change of base formula to express ⁵log 2 in terms of base 10 logarithms:

⁵log 2 = log 2 / log 5

We already know that log 2 ≈ 0.3010 and log 5 ≈ 0.6990. So,

⁵log 2 ≈ 0.3010 / 0.6990 ≈ 0.4306

Substitute this back into our expression:

(1/4) * (⁵log 2 + 1) ≈ (1/4) * (0.4306 + 1)

= (1/4) * 1.4306 ≈ 0.35765

So, the final answer for the third part is approximately 0.35765. This problem showed us how to handle fractional exponents and how to use a less common logarithm property to simplify the expression. It's all about having the right tools in your toolbox!

Final Thoughts and Tips for Logarithm Problems

Wow, we've tackled some pretty cool logarithm problems today! Remember, the key to mastering logarithms is understanding their properties and practicing applying them. Here are some final tips to keep in mind:

• Express numbers as powers of prime numbers: This often simplifies the logarithmic expressions.
• Use the power rule: log_b(m^p) = p * log_b(m) is your best friend!
• Change of base formula: Don't be afraid to switch to a more convenient base (like 10 or e).
• Product and quotient rules: These help you break down complex logarithms into simpler ones.
• Practice, practice, practice! The more you work with logarithms, the more comfortable you'll become.

Logarithms might seem tricky at first, but with a solid understanding of their properties and a bit of practice, you'll be solving these problems like a pro in no time. Keep up the great work, and remember to have fun with math! You've got this!