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.

We basically just need to learn 4 values and the rest we can calculate.

Now let’s apply to this problem

log is a monotonic function i.e. larger input values will give larger output values

So taking log of both sides

Left side:
7 * log 10 = 7

Right side:
10 * log 7 = 10 * .845 = 8.45

So clearly Right side is larger.

We can use this for larger bases by approximating them with numbers that have single digit factors. Like log 22 could be approximated to log 21 which is simply log 3 + log 7

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10**7 = 2**7 * 5**7

7**10 = 7**3 * 7**7

2**7 = 128

7**3 = 343

Since both components of 7**10 are larger, (B) follows.

Hi Chris

While I’m not a subscriber of Magoosh, I still value the articles here.

So I thought I’d share a few thoughts regarding this question

One way of solving such qns could be to use log with base 10.

This would require a bit of mugging up : Mainly we need to learn the log values from 1 to 10

(All logs below are base 10)

log 1 = 0

log 2 = .301

log 3 = .477

log 4 = 2 * log 2 (Since 2 = 2^2)

log 5 = .698

log 6 = log 2 + log 3 ( Since 6 = 2*3)

log 7 = .845

log 8 = 3 * log 2

log 9 = 3 * log 3

log 10 = 1

We basically just need to learn 4 values and the rest we can calculate.

Now let’s apply to this problem

log is a monotonic function i.e. larger input values will give larger output values

So taking log of both sides

Left side:

7 * log 10 = 7

Right side:

10 * log 7 = 10 * .845 = 8.45

So clearly Right side is larger.

We can use this for larger bases by approximating them with numbers that have single digit factors. Like log 22 could be approximated to log 21 which is simply log 3 + log 7

It’s really a nice explanation

Thank you!