Exponents are troublesome; fortunately, they do not need to be. What I plan to do in this series is show you the basics, so that you can move confidently through the intermediate level, and even the advanced level.

I’ve created three lessons, corresponding to basic, medium, and advanced. You can watch them here:

Once you have a sense of how the fundamentals of exponents work for each level, you can test yourself with a GRE problem. After all, knowing only the fundamentals does not mean you will answer a GRE question correctly.

Remember, the GRE is testing the way you think. That means the GRE will try to throw you off balance by presenting an exponents question that, at first glance, will give you pause. They expect you to be able to find an effective approach to the problem. Only then can you apply the fundamentals you’ve learned. In essence, the GRE is wrapping up a problem; unraveling a problem, so to speak, is half the battle.

**Level 1**

So now, I am going to take the fundamentals covered above, and I am going to wrap them up in a GRE problem. See if you can unravel the question below.

Practice Problem:

If , what is the value of y?

(A) 6

(B) 12

(C) 18

(D) 36

(E) 72

Answer: C

If you are even a bit unsure about any of the above, you will definitely want to watch the video.

**Level 2**

It’s going to be a little more difficult here, but if you are confident with level one, give Level 2 a try:

** **

Once again, I am going to take the fundamentals you’ve learned this far and I am going to turn them into a GRE problem:

** **

*, what is the value of n?*

*(A) **-2*

*(B) **0*

*(C) **1*

*(D) **2*

*(E) **4*

** **

Answer: D

If you were unable to answer the problem correctly, or if any of the above did not make sense, take a look at the video.

**Level 3**

Congratulations! You’ve already learned enough to answer most GRE exponent problems. In this level, you are going to learn more advanced techniques, perfect for those students looking to score above the top 85%.

Practice Problem:

If then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

*(A) **2*

*(B) **3*

*(C) **5*

*(D) 7*

*(E) **9*

Answer (D). This question is about as difficult as any exponent problem you will see on the GRE. So, if you got it right, good job! If you are unsure about the solution to the problem, you can see the steps in the picture below (scroll down):

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If you would like more practice on exponents, check out the Official Guide problems (there are a few exponent problems in the book). And, if you encounter any obstacles, don’t despair—I’ve recorded video explanations for every single problem in the Official Guide. For even more practice on exponents, and video explanations, give Magoosh a try.

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I really have no problem with the procedure, but the wording of the problem is really making me confused

Hi Moe,

The wording on the GRE is certainly difficult! They create questions that are meant to confuse you. If you let me know which question you are asking about, I might be able to help a little bit 🙂

For question number 3, I took the following approach to arrive at the answer:

n ={2^{-10}+2^{-9}}/{(7^{-1}) (5^9)}

= [ (1/2^ {10}) + (1/2^ {9}) ] / [(5^{9}/7)]

= [ (1/2^ {10}) + (1/2^ {9}) ] = [2^9 (1 + 10)] / (2^19) = 11 / (2^10)

= [11 / (2^10)] * [(7/5^{9})] = 77 / [(2^9)(2^1)(5^9)] = 77 / (10^9)(2^1)

=== (10^9) = 9 zeroes in the denominator. To divide a 2-digit number in the numerator by a number with 9 zeros in the denominator, move the decimal to the left 9 positions, leaving a decimal with 7 zeros.

Hi Lidiya,

I think you made one small calculation error here:

= [ (1/2^ {10}) + (1/2^ {9}) ] = [2^9 (1 + 10)] / (2^19)

In my calculations, I would have gotten the following:

= [ (1/2^ {10}) + (1/2^ {9}) ] = [2^9 + 2^10] / (2^19) = [2^9 (1 + 2)] / (2^19)

I hope this helps! 😀

Yes. Thank you! And that would lead to 21/[(2^10)(5^9)]

Does anyone have the solution to the last problem in level 3:

5^24 x 4^12

_______________________________ = ?

10^24

Suhail, it is as follows:

(5^24)(4^12) (5^24)({2^2}^12) (10^24)

————— = ——————– = ———– = 1

(10^24) (10^24) (10^24)

*** The question makes use of the rule : (A^x)(B^x) = ({AB}^x)

Where is the video for level 3, question 3? Thanks!

Hi Billy,

This is strange–it seems like the video has disappeared! I will let our blog writing team know about it so that they can fix the issue 🙂

In the meantime, you can scroll down through the other comments to get an explanation for the third question! Sorry about the confusion 🙂

For the level 3 problem, is it incorrect to move (2^-10 + 2^-9) from the numerator to the denominator? This is how I originally executed the problem. However, my result was 7/3(10^9) rather than 10.5/(10^9). I am assuming that (2^-10 + 2^-9) must first be factored before it can be moved. Why is this so?

Think of it this way,

2^-2 + 2^-1, is it equal to 1/ (2^2 + 2^1) ?

No,

2^-2 + 2^-1 = 1/4+ 1/2 = 3/4 where as, 1/ (2^2 + 2^1) = 1/(4+2) = 1/6

You have to factor out the common term with same exponent, then you can put the factor in the denominator with the opposite sign exponent.

Hi ,

I have problem with the last part of question.

Please help me understand how (10^-9)(7^1)(1.5) which can be written as 10.5 (10^-9) has 9 zero after decimal point.I think it should has 6 zero like 0.000000105

thanks

I do not understand how bringing the 2^-9 to the denominator gives us 10^-9, can someone please help? Also is there a video to show how to solve these questions because the lesson videos dont show how to do it.

Hi William,

Yeah, this is a pretty difficult question, and I can assure you the exponent related questions on the GRE won’t be this tough.

For this problem, you can multiply both the numerator and denominator by 2^9. Also, the answer should be 7, not 8. I’ve updated the question to reflect this. Sorry for any confusion :).

Hi Chris,

In problem 3, aren’t you counting the zero between 1 and 5 or do we have to restrict ourselves to count zeros from decimal till 1?

Otherwise, I see 8 zeroes in 0.0000000105.

*say=stay

Hi Len,

Does this refer to the text in the post? I can’t seem to find ‘say’.

Why can we move down 2^-9 to the denominator and 7^-1 to the numerator and why does -9 stay negative and -1 becomes positive after you move them? (is this a certain principle that I’m forgetting)?

Hi Len,

2^-9 does not stay negative; it becomes positive. That way you get (5^9)(2^9) in the denominator.

Hope that clears things up :).

I am a little confused. I dont understand how do you go from 2^-9(2^-1 + 1)/(7^-1×5^9) to –> =10^-9x7x1.5

thanks

V

Hi Valeria,

We can bring the 2^-9 down to the denominator. Giving us 10^-9. We can bring the 7^-1 to the numerator.

The (2^-1 + 1) = 1.5

That leaves us with (10^-9)(7^1)(1.5)

Hope that helps!

Why can we move down 2^-9 to the denominator and 7^-1 to the numerator and why does -9 say negative and -1 becomes positive when you move them? (is this a certain principle that I’m missing)?

I’m a little confused… I got (D) on the second level, and felt very confident in my answer. Plus, you said in your October 28 comment that the answer is (D). However, below the problem, it says the answer is (C). Can you tell me which is correct – and if it’s (C), why? Thanks!

Hi, Claire

The answer’s D– it was just a typo in the post, so you were right! Sorry for the confusion.

Best,

Margarette

Hi!

The level 2 GRE-type problem doesn’t have the answer listed. Is it (D) 2 ?

Yes, it’s (D). This question is pretty typical of something you would see on the GRE. As long as you follow your rules and don’t rush through the problem (and end up picking (A)) you should be able to get it right.

Hi,

Can you elaborate on the answer to the 3rd question?

I have solved it to this stage

n=(2^-10 + 2^-9)/(7^-1×5^9)

=2^-9(2^-1 + 1)/(7^-1×5^9)

=10^-9x7x1.5

=10^-9 x 10.5

Now what is to be done ? 10.5 is a terminating decimal. If i keep it at this stage the answer is 9. But i was a bit confused about whether i need to take the 0 in 10.5 that occurs before the decimal. So i moved it to after the decimal

If i simplify it 1.05 x 10^-8 then should i count the 0 that is there in 1.05 ??

Please clarify.

Good question Shyam! And this is quite a tough question – definitely 700-level GMAT.

So, after multiplying out 10^-9 x 10.5, you get .0000000105. The question asks how many zeroes after the decimal, so you have to count the zero between the 1 and the 5. That gives us a total of eight zeroes.

Hope that helps!

Thanks for the clarification Chris…