Okay, here is a tricky quantitative comparison question. Let’s see if you can get it in less than two minutes!

Column A Column B

10^7 7^10

^{ }

^{ }

One way of solving this problem is to do the math. However, doing so is a time-consuming, error-prone process, and though it may work, the GRE is not testing your ability to do laborious mathematics. Rather, it is testing the way you think, your logical approach to a complex problem.

One useful approach is to approximate. For instance if we take 7^2^{ }we get 49. Notice 49 is very close to 50. So let’s say 7^2 is equal to 50. Therefore (7^2)^2^{ }is equal to 50^2 or 2500. 2500 can also be represented as 2.5 x 10^3. So when we take 2500^2^{ }or 7^8 we can convert 2500 into 2.5 x 10^3^{ }and we have (2.5 x 10^3)^2. Again, because we are approximating let’s round 2.5 down to 2. Now we have (2 x 10^3)^2 = 4 x 10^6 = 7^8 . This figure is almost as large as 10^7. Since we are comparing 7^10^{ }to 10^7 you can see that 7^10 will be much larger.

Granted, this method still involves a decent amount of calculation. Another technique we can apply is pattern recognition. For example, we can use smaller numbers that adhere to the pattern above. So lets compare 2^5^{ }to 5^2. Notice that I’ve kept a similar pattern to the original question: x^y^{ }vs. y^x^{. }, where y = x + 3. In this case we get 2^5_{ }= 32 and 5^2 = 25. Notice that the number with the greater exponent is greater. We should try one more set of numbers just to be sure, 3^6^{ }vs. 6^3. Notice that this time the difference between 3^6^{ }and 6^3^{ }is even greater than that between 2^5 and 5^2. That is, the difference between x^y and y^x^{ }where y = x + 3 increases the larger the numbers we plug in. Therefore 7^10^{ }will be much greater than 10^7. The answer is (B).

This question is by no means an easy one. In fact it is probably at the 700-level. The key is to not spend three or four minutes calculating both columns. Remember, when you see huge numbers or ones that seem very difficult to calculate come up with a logical approach.

For the next part of this blog post, I’ll provide a similar problem. Try to use the logical approach vs. solving approach.