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12 Oct 2022, 08:31
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Playing with Powers
Let us start this article with a very interesting question
If \(a\) is a real number, is \(a^3>a\)?
How do I know for what values is \(a^3>a\)?
numberlineGC.png [ 5.45 KiB | Viewed 2106 times ]
More about Number line
numberlineGC2.png [ 5.58 KiB | Viewed 2113 times ]
\(1^{st}\) Region
numberlineGC3.png [ 2.46 KiB | Viewed 2118 times ]
\(2^{nd}\) Region
numberlineGC4.png [ 2.45 KiB | Viewed 2115 times ]
\(3^{rd}\) Region
numberlineGC5.png [ 2.47 KiB | Viewed 2042 times ]
\(4^{th}\) Region
numberlineGC6.png [ 2.43 KiB | Viewed 2070 times ]
Let us now go back to the question we started with
So, we can conclude that the only region where this inequality holds true are \(1<a<+∞\) and \(-1<a<0\)
What about terminal points?
Now you have noticed that we have talked about regions \(1<a<+∞\), \(0<a<1\), \(-1<a<0\), and \(-∞ < a < -1\)
Let us start this article with a very interesting question
If \(a\) is a real number, is \(a^3>a\)?
- You might say yes, it is because \(2^3>2\), \(3^3>3\) and so on
- However, what about if \(a=-2\)?
- \((-2)^3=-8\) and we can no longer say that \(a^3>a\)
- This is where powers/exponents/indices become interesting
How do I know for what values is \(a^3>a\)?
- The answer to this question lies in the old-age concept of Number Line
- We know what a number line looks like, right? Primarily, it has 2 regions
- On one side we have all the positive numbers leading to infinity "\(∞\)"
- On the other side we have all the negative numbers leading to negative infinity "\(-∞\)"
- Also, there is a point separating these two regions, zero (\(0\))
- Do keep in mind it is neither negative or nor positive
Attachment:
numberlineGC.png [ 5.45 KiB | Viewed 2106 times ]
More about Number line
- However, to tackle the questions we asked, we need to further split this line
- In GMAT, we need to consider not 2 but 4 regions
- And this is how it looks like -
Attachment:
numberlineGC2.png [ 5.58 KiB | Viewed 2113 times ]
- Let’s talk about each region one by one
\(1^{st}\) Region
- The 1st region of a number line is beyond 1 till infinity
Attachment:
numberlineGC3.png [ 2.46 KiB | Viewed 2118 times ]
- To understand how exponents behave in this region, let’s take an example of \(a=2\)
- If \(a=2\), then
- \(a^2=4>a\)
- \(a^3=8>a^2\)
- \(a^4=16>a^3\)
- If we use radical power over \(a=2\), then
- \(a^{\frac{1}{2}}=1.414<a\)
- \(a^{\frac{1}{3}}=1.259<a^{\frac{1}{2}}\)
- If \(a=2\), then
- Please note that not only 2 but any number in this region will follow the same trend
- So, the trend that we see in this region is:
\(a<a^2<a^3<a^4<⋯⋯⋯\) and \(a>a^{\frac{1}{2}}>a^{\frac{1}{3}}>a^{\frac{1}{4}}>⋯⋯⋯\)
- Note: memorizing these trends are not necessary. We can always take an easy example like we did and get the trend instantly
\(2^{nd}\) Region
- The 2nd region of a number line is between 0 and 1
Attachment:
numberlineGC4.png [ 2.45 KiB | Viewed 2115 times ]
- Let’s take an example of \(a=0.2\)
- If \(a=0.2\), then
- \(a^2=0.04<a\)
- \(a^3=0.008<a^2\)
- \(a^4=0.0016<a^3\)
- If we use radical power over \(a=0.2\), then
- \(a^{\frac{1}{2}}=0.447>a\)
- \(a^{\frac{1}{3}}=0.584>a^{\frac{1}{2}}\)
- If \(a=0.2\), then
- So, the trend that we see in this region is like:
\(a>a^2>a^3>a^4>⋯⋯⋯\) and \(a<a^{\frac{1}{2}}<a^{\frac{1}{3}}<a^{\frac{1}{4}}<⋯⋯⋯\)
\(3^{rd}\) Region
- The 3rd region of a number line is between -1 and 0
Attachment:
numberlineGC5.png [ 2.47 KiB | Viewed 2042 times ]
- Let’s take an example of \(a=-0.2\)
- Note: since this is negative, the even powers will always be greater than the odd powers. Therefore, let us compare even and odd powers separately
- If \(a=-0.2\), then
- \(a^3=-0.008>a\)
- \(a^5=-0.00032>a^3\)
- \(a^2=0.04\)
- \(a^4=0.0016<a^2\)
- \(a^6<a^4\)
- So, the trend that we see in this region is like:
\(a<a^3<a^5<a^5<⋯⋯⋯\) and \(a^2>a^4>a^6>a^8>⋯⋯⋯\)
\(4^{th}\) Region
- The 4th region of a number line is less than -1 till negative infinity
Attachment:
numberlineGC6.png [ 2.43 KiB | Viewed 2070 times ]
- We would like you to take the example of \(a=-1.2\) and find the trend for this region
- The trend that we see in this region is like:
\(a>a^3>a^5>a^5>⋯⋯⋯\) and \(a^2<a^4<a^6<a^8<⋯⋯⋯\)
Let us now go back to the question we started with
- \(a^3>a\) is applicable only in some of these regions
- Ideally, we must check for these 4 lines. Let us understand how we check it
- 1st region, \(1<a<+∞\)
- We can take \(a = 2\) here and see \(2^3>2\)
- Hence, the first region is valid region for the given inequality
- 2nd region, \(0<a<1\)
- We can similarly take \(a=\frac{1}{2}\)
- Here, \(\frac{1}{2}^3 <\frac{1}{2}\)
- This region is not valid for the given inequality
- 3rd region, \(-1<a<0\)
- We can similarly take \(a=-\frac{1}{2}\)
- Here, \((-\frac{1}{2})^3 >\frac{1}{2}\)
- This region is valid for the given inequality
- 4th region, \(-∞ < a < -1\)
- We can take a = - 2 here and see \((-2)^3< -2\)
- Hence, this region is not valid for the given inequality
- 1st region, \(1<a<+∞\)
So, we can conclude that the only region where this inequality holds true are \(1<a<+∞\) and \(-1<a<0\)
What about terminal points?
Now you have noticed that we have talked about regions \(1<a<+∞\), \(0<a<1\), \(-1<a<0\), and \(-∞ < a < -1\)
- Do you think we have discussed all the possible cases?
- No, we haven’t. We are still left with the terminal points i.e., scenarios when \(a=-1,0\) or \(1\)
- These cases should always be dealt with separately and should not be missed
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13 Oct 2022, 06:38
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16 Oct 2022, 23:34
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The questions are posted now and you can find the link in the above post.
Solutions will be posted after 48 hours.
Solutions will be posted after 48 hours.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
