Mathematical Proof: Solving The Exponential Equation
Hey guys! Let's dive into a cool math problem! We're gonna break down how to solve an exponential equation step-by-step. The goal is to show that a certain expression is a whole number. This problem is a great example of how mathematical concepts like exponents, logarithms (though we won't explicitly use them here), and algebraic manipulation come together. Get ready to flex those brain muscles! So, the problem is: If 36a = 4 and 36b = 3, show that 25(1-a-b)/(1-a) ∈ Z (which means the result is an integer). Let's start with breaking down the given information and finding a path to get to the answer. It might look a little intimidating at first, but trust me, with careful steps, we can crack it!
Unpacking the Problem: Understanding the Basics
Alright, so what do we actually know from the start, you ask? Well, we have two key equations: 36a = 4 and 36b = 3. These tell us how the number 36 behaves when raised to powers 'a' and 'b'. The core idea here is to manipulate these equations to build towards the expression we're trying to prove is an integer. Remember that the power/exponent tells us how many times a number is multiplied by itself. It's crucial to realize that numbers can be expressed in terms of other numbers using their exponents. For example, the number 36 itself can be broken down. Since 36 is the result of 6*6, so 36 can be written as 62. Similarly, 4 is also a power of 2 (22), and 3 is the square root of 9. So we are looking at numbers that have relationships between each other, allowing us to find new insights! The most important strategy is to use the given facts and build the expression. We can express the numbers with exponents. Since we have both a and b in the final expression, this implies that we will use both of the equations to find the result. And finally, remember that our goal is to show that the final result is an integer, so we need to deal with the equations with a logical approach. We can take this step by step, guys! Don't you worry about that!
Manipulating the Equations: Getting Closer to the Solution
Okay, let's start getting our hands dirty and actually manipulate those equations. From 36a = 4, we can express 'a' in terms of logarithms, but as I said, we won't explicitly use logarithms here; instead, we'll try to find a direct path. What's important is that we notice 36 and 4 are both powers of 6 and 2, and we can find a relationship between the two. From the second equation, 36b = 3, we see that 'b' relates to the number 3. The numbers 3, 4, 36, and 25 seem random, but we will make them talk to each other to solve the problem. The expression that we need to find is 25(1-a-b)/(1-a). Let's see if we can manipulate the known equations to get closer to what we need. Let's see where that takes us. The equation has 1-a-b in the numerator and 1-a in the denominator. That means we should start with either 'a' or 'b' from the two equations. Remember, the goal is to get to the final expression. We will use the two equations, so let's start with a from the first equation, 36a = 4.
Since 36a = 4, we can raise both sides to a power to try to get closer to our target expression. Let's think about it: we want (1-a-b) and (1-a) in the expression. We can try to build an expression for (1-a). Let's divide both sides by 36, which is 361.
So, 36a / 361 = 4 / 36
Using exponent rules, we get 36a-1 = 1/9
We also know that 36b = 3, and we will try to make use of this.
Connecting the Pieces: The Road to the Final Answer
Now, how do we involve b in this? We also know 36b = 3. Our goal is to connect all these pieces and manipulate them to obtain the desired expression: 25(1-a-b)/(1-a). The key is to notice the relationship between the numbers and their exponents. Since 36a-1 = 1/9 and 36b = 3, and since the target expression is 25(1-a-b)/(1-a), we have to look for numbers that relate to each other. Notice that 1/9 = 3-2. And we also know that 36 = 62, so we know that the relationship between 36 and 3 can also be applied to the equations. So we will take both equations and multiply them to get closer to our result. Also, we can see that we have 1-a in the exponent. So, we'll try to figure out how to generate the 1-a-b on the exponent. We can rewrite the first equation, 36a = 4, we can take the log on both sides to find a relationship between 36 and 4. But we won't take that path since we can relate 4 and 3 with the base 2 and 3. Now, let's consider the two equations again:
- 36a = 4
- 36b = 3
We want to get to 1-a-b. Let's try to manipulate them to see what we can find.
Multiplying the two equations, 36a * 36b = 4 * 3. Therefore, 36a+b = 12.
Now we want to include the 1. We also know that 361 = 36. So let's divide both sides by 36 again:
36 / 36a+b = 36/12
Using exponent rules, we get 361-a-b = 3
Now we are getting somewhere, guys! Remember our goal is to find the expression 25(1-a-b)/(1-a) and show it's an integer. Let's see what we have now.
The Grand Finale: Proving the Integer
Okay, guys! We've made great progress. We have 361-a-b = 3. Previously, we also found that 36a-1 = 1/9. Now, let's use these two results. Since we know that 361-a-b = 3, we can raise both sides to some power to get the same form with the expression. We can see that from the previous step, we can get the following expression.
We know 361-a-b = 3. Now let's try to relate this to 36a-1 = 1/9.
Also, we know that 36a-1 = 1/9 = 3-2
So we can replace 3 with 361-a-b. Also, we can express 3 as (361-a-b)-2
Therefore, we have (361-a-b)-2 = 36a-1. Using exponent rules, we can rewrite the equation as 36(1-a-b)*(-2) = 36a-1. Hence, we can match the exponents and write the following equation.
(1-a-b) * (-2) = a - 1.
(-2 + 2a + 2b) = a - 1.
2a + 2b - a = -1 + 2
a + 2b = 1
Using the equations, we will try to transform to find out if 25(1-a-b)/(1-a) ∈ Z. We can see that 361-a-b = 3. We can write 36 as 62, which means (62)1-a-b = 3. Using exponent rules, we have 62(1-a-b) = 3. Therefore, 6(1-a-b) = 31/2. Let's go back to our main goal and remember that we have to transform 25, since 25 is also a number. Since we have a lot of 3s, let's start with 25(1-a-b)/(1-a). Also, we know that 25 is also a power of 5, so we can write this equation as (52)(1-a-b)/(1-a). Then we have 52*(1-a-b)/(1-a). Since we got to 3, let's also take a look at the previous results to express the form. We know that 36b = 3 and 361-a-b = 3. Then, we can find that 361-a-b = 36b. Then 1-a-b = b. So, 1 - a = 2b. From the formula (52)(1-a-b)/(1-a), we have 52*(1-a-b)/(1-a). From the previous step, we have 1 - a = 2b, so we can replace the expression, which is 52*(1-a-b)/(2b). Then, we know that (1-a-b) = b, so we can rewrite the equation to 52*b/2b = 51. Therefore, the result of 25(1-a-b)/(1-a) is 5. Since 5 is an integer, we have finally proved that 25(1-a-b)/(1-a) ∈ Z. That's a wrap!
Therefore, 25(1-a-b)/(1-a) = 5, which is an integer. Thus, 25(1-a-b)/(1-a) ∈ Z.