Plugging Numbers Without Using Transition Points
BY KARISHMA, VERITAS PREP
A few months back, one of our posts talked about knowing which numbers to plug-in in case you want to use the number–plugging method. To be more exact, we discussed that you need to find the transition points i.e. the points where the two sides of the inequality become equal. The transition points tend to reverse the relation between the two sides. For a detailed discussion of this concept, revisit this post.
A question that arises here is what if the transition points are not apparent? What do we do in that case? First of all, let us say that the use of logic is preferable in every question. There are few questions (but there sure are!) where the number plugging method is the only decent option. But that’s beside the point. Let’s take up a question and see what to do in case the transition point is hard to see.
Question: If x is an integer, is 4^x < 3^(x+1)?Statement I: x is positive
Statement II: |x – 1| < 2
Solution:Again, our moral duty is to first give you the logical solution since we would like you to think in those terms as far as possible. (Though in this question, plugging in numbers might seem easier.) We will discuss how to get the answer by plugging in appropriate numbers in this case.
Method 1: Logical SolutionThe question stem only tells us that x can take any integral value. We need to find whether \(4^x\) is less than \(3^{(x+1)}\).
Consider statement I: x is positive. Given that x is positive, is \(4^x\) less than \(3^{(x+1)}\)? If we put x = 1, it is easy to see that 4^x is less than 3^(x+1). So the inequality holds in this case. The point is how do we prove that this will be true for all positive values of x? It’s tough to prove that something holds for a lot of numbers. It’s easier to show that it doesn’t hold for at least one number since we need only one suitable value in that case.
Question Stem: Is \(4^x < 3^{(x+1)}\)?
Reframe it as: Is \(4^x < 3^x * 3\)?
Or Is \((\frac{4}{3})^x < 3\)?
Note that 4/3 (= 1.33) is greater than 1. When you raise it to a very high power, it will take a very large value. There is no reason it should stay less than 3. Hence the inequality will not hold for large values of x. Hence this statement alone is not sufficient.
Some properties to note:
* When you raise a positive number greater than 1 to a large positive integer power, it takes a large value.
* When you raise 1 to a large positive integer power, it stays 1.
* When you raise a positive number less than 1 to a large positive integer power, the number becomes even smaller than the original value.
These are some number properties you need to work through and be comfortable with.
Consider statement 2: \(|x – 1| < 2\)
Hopefully, you understand modulus well now. We can say that this inequality implies that x is a point at a distance less than 2 from the point 1 on the number line i.e. -1 < x < 3.
Is \((\frac{4}{3})^x < 3\)?
For small values of x e.g. x = 0, we know the inequality holds. Let’s check for only the largest value x can take i.e. 2 since x must be an integer. Even if x were 2, \((\frac{4}{3})^x = \frac{16}{9}\) i.e. less than 2. \((\frac{4}{3})^x\) would still be less than 3. Hence the inequality will hold in this case. This statement alone is sufficient to answer the question.
Note that we did use some basic number plugging here too but that number plugging helps us get a clear picture and makes us ask the right questions. There is nothing wrong with plugging in some numbers here and there to understand the logic. If you know why you are plugging in a particular number, it means you are on the right track. Blindly plugging in is the problem.
2. Putting in NumbersLet’s look at this method too now.
Is \(4^x < 3^{(x+1)}\)?
Here it’s not easy to find the transition point. We would have to plug in numbers again to find where the two sides of the inequality are equal! So let’s ignore the transition points and directly start plugging in numbers.
Statement 1: x is positive.
Put x = 1, you get 4 < 9 (Holds)
Put x = 2, you get 16 < 27 (Holds)
Put x = 3, you get 64 < 81 (Holds)
What do we do now? How long are we supposed to keep putting in numbers? We cannot do it for all positive integers. How do we decide when to stop? Note that the relative difference between the left hand side and the right hand side is reducing. 9 is more than twice of 4. 27 is more than 16 but not quite twice. It is more than 1.5 times of 16. 81 is somewhat more than 64 but not more than 1.5 times of 64. What this means is that soon enough, the difference will go to 0 and left hand side will become more than the right hand side. If you want to check and you are comfortable with higher powers of numbers,
Put x = 4, you get 256 < 243 (Does not hold)
What you need to do in case the transition point is not apparent is focus on the pattern of the numbers. Is the difference between them narrowing or widening?
This statement alone is not sufficient to answer the question.
Statement II: |x – 1| < 2
As above, we get -1 < x < 3 so this is simply a matter of putting in x = 0, 1 and 2 to see that the inequality holds in each case. Sufficient.
Answer (B) This question is discussed HERE.