Solving Logarithmic Expressions: A Step-by-Step Guide

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Hey math enthusiasts! Let's dive into a cool logarithmic problem. We're going to break down how to solve the expression (2log⁑10)(5log⁑10)βˆ’(2log⁑5+5log⁑2)(^2\log 10)(^5\log 10) - (^2\log 5 + ^5\log 2). This might look a little intimidating at first, but trust me, we'll go through it step by step, making it super easy to understand. Ready to get started?

Understanding the Basics of Logarithms

Before we jump into the problem, let's quickly recap some fundamental concepts about logarithms. Think of a logarithm as the inverse operation of exponentiation. If we have an expression like blog⁑a=c^b\log a = c, this is equivalent to saying bc=ab^c = a. Here, b is the base of the logarithm, a is the argument (the number we're taking the logarithm of), and c is the exponent. Understanding this relationship is key to solving logarithmic problems.

Also, remember these essential rules of logarithms:

  1. Change of Base Formula: This is super useful! It lets you change the base of a logarithm: alog⁑b=clog⁑bclog⁑a^a\log b = \frac{^c\log b}{^c\log a}.
  2. Product Rule: The logarithm of a product is the sum of the logarithms: blog⁑(xy)=blog⁑x+blog⁑y^b\log(xy) = ^b\log x + ^b\log y.
  3. Quotient Rule: The logarithm of a quotient is the difference of the logarithms: blog⁑(xy)=blog⁑xβˆ’blog⁑y^b\log(\frac{x}{y}) = ^b\log x - ^b\log y.
  4. Power Rule: The logarithm of a number raised to a power is the power times the logarithm of the number: blog⁑(xn)=nβ‹…blog⁑x^b\log(x^n) = n \cdot ^b\log x.

These rules are like secret weapons in the world of logarithms, making complex problems much simpler. They help us manipulate logarithmic expressions, simplify them, and eventually solve for the unknown. Keep these in mind as we work through our main problem.

Now, let's put these rules into action! We'll use them to break down the given expression and find its value. Remember, the goal is to transform the expression into a form we can easily evaluate. With each step, we'll simplify and apply the appropriate logarithmic rules to get closer to our solution. So, let’s get into the main dish of our discussion!

Breaking Down the Expression: (2log⁑10)(5log⁑10)(^2\log 10)(^5\log 10)

Alright, let’s tackle the first part of our expression: (2log⁑10)(5log⁑10)(^2\log 10)(^5\log 10). Our aim is to simplify this part as much as possible. Notice that we have two logarithmic terms multiplied together. This is where we can apply a clever trick using the change of base formula. Remember that formula? It's our best friend here.

Let's start by rewriting each logarithmic term using the change of base formula. We'll choose a common base, like 10, to make things simpler. So, we'll rewrite 2log⁑10^2\log 10 and 5log⁑10^5\log 10 using base 10.

2log⁑10=10log⁑1010log⁑2=110log⁑2^2\log 10 = \frac{^{10}\log 10}{^{10}\log 2} = \frac{1}{^{10}\log 2} And

5log⁑10=10log⁑1010log⁑5=110log⁑5^5\log 10 = \frac{^{10}\log 10}{^{10}\log 5} = \frac{1}{^{10}\log 5}

Now, substitute these back into the original expression:

(2log⁑10)(5log⁑10)=(110log⁑2)(110log⁑5)(^2\log 10)(^5\log 10) = (\frac{1}{^{10}\log 2})(\frac{1}{^{10}\log 5})

This gives us

1(10log⁑2)(10log⁑5)\frac{1}{(^{10}\log 2)(^{10}\log 5)}

However, this form does not seem to easily simplify further. Let’s take another approach. Instead of changing to base 10 directly, let's use the property alog⁑b=1blog⁑a^a\log b = \frac{1}{^b\log a}. So,

2log⁑10=2log⁑(2βˆ—5)=2log⁑2+2log⁑5=1+2log⁑5^2\log 10 = ^2\log (2*5) = ^2\log 2 + ^2\log 5 = 1 + ^2\log 5

5log⁑10=5log⁑(2βˆ—5)=5log⁑2+5log⁑5=1+5log⁑2^5\log 10 = ^5\log (2*5) = ^5\log 2 + ^5\log 5 = 1 + ^5\log 2

(2log⁑10)(5log⁑10)=(1+2log⁑5)(1+5log⁑2)(^2\log 10)(^5\log 10) = (1 + ^2\log 5)(1 + ^5\log 2)

=1+5log⁑2+2log⁑5+(2log⁑5)(5log⁑2)= 1 + ^5\log 2 + ^2\log 5 + (^2\log 5)(^5\log 2).

Notice that if we can do something with the last term, then we are on the right track. Let’s save this and go to the next part.

Simplifying the Expression: (2log⁑5+5log⁑2)(^2\log 5 + ^5\log 2)

Next up, we need to simplify (2log⁑5+5log⁑2)(^2\log 5 + ^5\log 2). This part looks a bit more straightforward. We have two logarithmic terms added together. Unfortunately, we can't directly simplify this using a single logarithmic rule like the product or quotient rule, because the bases are different. However, we have a valuable trick up our sleeves:

  • The Change of Base Formula: This allows us to rewrite these logarithms with a common base. Although, using the common base does not simplify the problem much.

Let’s try to manipulate the terms. Notice the relationship between the two terms.

Recall that alog⁑b=1blog⁑a^a\log b = \frac{1}{^b\log a}.

So, we can rewrite 5log⁑2^5\log 2 as 12log⁑5\frac{1}{^2\log 5}.

2log⁑5+5log⁑2=2log⁑5+12log⁑5^2\log 5 + ^5\log 2 = ^2\log 5 + \frac{1}{^2\log 5}

This expression looks like it could be part of a larger, more complex equation, but on its own, it doesn't simplify nicely into a single number. So let us keep it like that for now.

Now, if we observe both parts, we can see that in the first part, we get 1+5log⁑2+2log⁑5+(2log⁑5)(5log⁑2)1 + ^5\log 2 + ^2\log 5 + (^2\log 5)(^5\log 2), and in the second part we have 2log⁑5+5log⁑2^2\log 5 + ^5\log 2. The second part already exists in the first part, so let us try to simplify the first part by changing the base.

Recall that

2log⁑10=2log⁑(2βˆ—5)=2log⁑2+2log⁑5=1+2log⁑5^2\log 10 = ^2\log (2*5) = ^2\log 2 + ^2\log 5 = 1 + ^2\log 5

5log⁑10=5log⁑(2βˆ—5)=5log⁑2+5log⁑5=1+5log⁑2^5\log 10 = ^5\log (2*5) = ^5\log 2 + ^5\log 5 = 1 + ^5\log 2

(2log⁑10)(5log⁑10)=(1+2log⁑5)(1+5log⁑2)(^2\log 10)(^5\log 10) = (1 + ^2\log 5)(1 + ^5\log 2)

=1+5log⁑2+2log⁑5+(2log⁑5)(5log⁑2)= 1 + ^5\log 2 + ^2\log 5 + (^2\log 5)(^5\log 2).

Let us consider 2log⁑5βˆ—5log⁑2^2\log 5 * ^5\log 2.

Using change of base formula:

2log⁑5βˆ—5log⁑2=10log⁑510log⁑2βˆ—10log⁑210log⁑5=1^2\log 5 * ^5\log 2 = \frac{^{10}\log 5}{^{10}\log 2} * \frac{^{10}\log 2}{^{10}\log 5} = 1.

So, (2log⁑10)(5log⁑10)=1+5log⁑2+2log⁑5+1=2+5log⁑2+2log⁑5(^2\log 10)(^5\log 10) = 1 + ^5\log 2 + ^2\log 5 + 1 = 2 + ^5\log 2 + ^2\log 5.

Putting It All Together: Final Calculation

Okay, guys, we have all the pieces now! Let's put everything back together and find the final answer. Remember, our original problem was:

(2log⁑10)(5log⁑10)βˆ’(2log⁑5+5log⁑2)(^2\log 10)(^5\log 10) - (^2\log 5 + ^5\log 2)

From our previous calculations, we know that:

  • (2log⁑10)(5log⁑10)=2+5log⁑2+2log⁑5(^2\log 10)(^5\log 10) = 2 + ^5\log 2 + ^2\log 5
  • 2log⁑5+5log⁑2=2log⁑5+5log⁑2^2\log 5 + ^5\log 2 = ^2\log 5 + ^5\log 2

Substituting these values into the original expression, we get:

(2+5log⁑2+2log⁑5)βˆ’(2log⁑5+5log⁑2)(2 + ^5\log 2 + ^2\log 5) - (^2\log 5 + ^5\log 2)

=2+5log⁑2+2log⁑5βˆ’2log⁑5βˆ’5log⁑2= 2 + ^5\log 2 + ^2\log 5 - ^2\log 5 - ^5\log 2

$= 2

So, the answer is 2! Awesome work, everyone!

Conclusion: The Answer Revealed!

And there you have it, folks! We've successfully solved the logarithmic expression. By breaking it down step-by-step, applying the right logarithmic rules, and a little bit of algebraic manipulation, we were able to simplify the expression and find the answer. Remember, practice is key when it comes to math. The more you work with these concepts, the easier they'll become. So, keep practicing, and you'll become a logarithmic master in no time!

So, the answer to our original problem (2log⁑10)(5log⁑10)βˆ’(2log⁑5+5log⁑2)(^2\log 10)(^5\log 10) - (^2\log 5 + ^5\log 2) is B. 2. Congratulations!