Solving ³log 160 - ³log 45 - ³log 16 - ³log 6: A Step-by-Step Guide

Hey guys! Ever stumbled upon a logarithmic expression that looks like it came straight out of a math textbook and made you scratch your head? Well, today, we're diving deep into one such expression: ³log 160 - ³log 45 - ³log 16 - ³log 6. Don't worry, it's not as intimidating as it looks! We'll break it down step by step, making sure you understand not just the solution, but the why behind it. So, buckle up, and let's get started on this logarithmic adventure!

Breaking Down the Basics of Logarithms

Before we jump into solving this particular problem, let's quickly recap the fundamental concepts of logarithms. Think of logarithms as the inverse of exponentiation. If you have an expression like b^y = x, the logarithmic form is written as log_b(x) = y. Here, 'b' is the base, 'x' is the argument, and 'y' is the exponent. Essentially, the logarithm answers the question: “To what power must we raise the base 'b' to get 'x'?”

In our case, we're dealing with logarithms to the base 3 (³log). This means we're looking for the power to which we must raise 3 to get the number inside the logarithm. Understanding this basic principle is crucial for manipulating and simplifying logarithmic expressions. Remember, logarithms are just a different way of expressing exponents, and they come with their own set of rules that make calculations easier.

Key Logarithmic Properties to Remember

To effectively tackle the expression ³log 160 - ³log 45 - ³log 16 - ³log 6, we need to have a firm grasp on some key logarithmic properties. These properties are like the secret ingredients in our math recipe, allowing us to combine and simplify terms. Here are the ones we'll be using today:

1. Product Rule: log_b(mn) = log_b(m) + log_b(n). This rule tells us that the logarithm of a product is equal to the sum of the logarithms of the individual factors.
2. Quotient Rule: log_b(m/n) = log_b(m) - log_b(n). This is the flip side of the product rule. The logarithm of a quotient is equal to the difference between the logarithms of the numerator and the denominator.
3. Power Rule: log_b(m^p) = p * log_b(m). This rule allows us to bring exponents outside of the logarithm as coefficients.

These properties are essential tools in simplifying logarithmic expressions. By understanding how to apply them, we can transform complex problems into more manageable ones. In our example, we'll primarily be using the quotient rule to combine the logarithmic terms.

Applying Logarithmic Properties to the Problem

Now, let’s get our hands dirty and apply these properties to our expression: ³log 160 - ³log 45 - ³log 16 - ³log 6. The first thing we notice is that we have a series of subtractions. This is where the quotient rule comes into play. Remember, the quotient rule states that log_b(m/n) = log_b(m) - log_b(n). We can use this rule to combine the terms.

Let's rewrite the expression by combining the subtractions into divisions within a single logarithm. This is like reversing the quotient rule. We'll group the terms being subtracted and rewrite them as divisors:

³log 160 - ³log 45 - ³log 16 - ³log 6 = ³log (160 / 45) - ³log 16 - ³log 6

Now, we can combine the next subtraction:

³log (160 / 45) - ³log 16 = ³log [(160 / 45) / 16]

And finally, the last one:

³log [(160 / 45) / 16] - ³log 6 = ³log {[(160 / 45) / 16] / 6}

So, our expression now looks like this: ³log {[(160 / 45) / 16] / 6}. We've successfully condensed the multiple logarithmic terms into a single logarithm. This is a huge step forward!

Simplifying the Expression Inside the Logarithm

Okay, we've managed to consolidate our logarithmic expression into a single term, which is excellent progress. But, the argument inside the logarithm, {[(160 / 45) / 16] / 6}, looks a bit messy, doesn't it? Our next task is to simplify this fraction. Remember, when you divide by a fraction, it's the same as multiplying by its reciprocal. So, let's rewrite the divisions as multiplications:

{[(160 / 45) / 16] / 6} = (160 / 45) * (1 / 16) * (1 / 6)

Now, we can simplify this by canceling out common factors. Let's break down the numbers into their prime factors to make this easier:

• 160 = 2^5 * 5
• 45 = 3^2 * 5
• 16 = 2^4
• 6 = 2 * 3

Substituting these into our expression, we get:

(2^5 * 5) / (3^2 * 5) * (1 / 2^4) * (1 / (2 * 3))

Now, let's cancel out the common factors:

• The '5' in the numerator and denominator cancels out.
• 2^5 in the numerator and 2^4 * 2 (which is 2^5) in the denominator cancels out.

This leaves us with:

1 / (3^2 * 3) = 1 / (3^3) = 1 / 27

So, we've simplified the expression inside the logarithm to 1/27. Our expression now looks much cleaner: ³log (1/27).

Calculating the Final Value

We've arrived at the final stage! We now have the simplified expression ³log (1/27). Remember what a logarithm means? It's the power to which we must raise the base (in this case, 3) to get the argument (1/27). So, we're asking ourselves: “To what power must we raise 3 to get 1/27?”

We know that 27 is 3 cubed (3^3). And, a fraction with 1 in the numerator and a power of 3 in the denominator can be expressed as a negative exponent. In other words:

1/27 = 1/(3^3) = 3^(-3)

Therefore, ³log (1/27) = ³log (3^(-3)).

Now, it's clear that the power to which we must raise 3 to get 3^(-3) is -3. So, the final answer is:

³log (1/27) = -3

Conclusion: Mastering Logarithmic Expressions

And there you have it! We've successfully navigated the logarithmic expression ³log 160 - ³log 45 - ³log 16 - ³log 6 and arrived at the solution: -3. We did this by understanding the fundamental properties of logarithms, applying the quotient rule to combine terms, simplifying the expression inside the logarithm, and finally, calculating the value.

The key takeaway here is that logarithmic problems, while they may seem daunting at first, can be broken down into manageable steps. By mastering the basic properties and practicing regularly, you can become confident in your ability to tackle any logarithmic challenge. So, keep practicing, keep exploring, and remember that math can be an exciting journey of discovery! You've got this!

Keywords: Logarithms, Logarithmic Properties, Quotient Rule, Simplifying Expressions, Mathematics, Solving Logarithms, Base 3 Logarithms