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# Math Basics: Exponents

One thing is certain on test day – you will see exponents. Another thing is also certain – those exponent problems will be meant to trick you. So, know your exponent properties, but also be aware that knowing the fundamentals alone won’t guarantee success. Tread carefully, and check your work. You should be able to answer most of these exponent questions correctly. Speaking of which, let’s see if you can answer the following three questions. They are not very difficult, only a little tricky.

1. 1. 2. 3. 4. 5. 2. 1. 2. 3. 4. 5. 3. If , what is the value of ?

1. 2. 3. 4. 5. For the first problem, do not simply add the exponents together. You can only do that when you are multiplying exponents, not adding them. Instead, notice how there are 4 of the same number, . Therefore, . Answer (C).

The next problem also asks you to add exponents, but first, you must multiply the exponents in parenthesis to the exponent outside the parenthesis, giving you: . Remember, that does not add up to . If you factor from both and , you get: . Answer D. Also, you can use approximation and eliminate those answers that are too small and too big. Answer choice E is far too big, and Answer choice A too small.

For the last problem, the best way to deal with it is by plugging in the answer choices. B, C, and D will all result in some funky irrational number because you are taking the square or cube root of the number. That leaves us with (A) -1 and (E) 1. Plugging in (A), you end up getting . Therefore, the answer has to be (E) 1. Remember, also, that any number taken to a negative exponent results in the reciprocal of that number, e.g. . And .

If you need more practice with exponents, look no further than Magoosh – both our products (oh, yeah, did I mention – we just released our new GRE product, over 600 practice questions waiting for you!) and on previous blog posts. Good luck!

P.S. Ready to improve your GRE score? Get started today.

### 25 Responses to Math Basics: Exponents

1. GPG February 23, 2018 at 5:15 am #

Hi Chris,
For question 2, I made the exponents same=2 and bases as 2^10 and 2^3 which could be multiplied to finally give me answer of 2^26.
Detailed step-by-step
(2^5)^4 + (2^3)^2
=(2^10)^2 + (2^3)^2
= (2^10 * 2^3)^2
=(2^13)^2
=2^26

What could have gone wrong here?

• Magoosh Test Prep Expert February 23, 2018 at 12:05 pm #

Hi there,

I think I see what happened here 🙂 You were multiplying the two exponents together, but that isn’t a valid operation here. The exponent outside the parentheses applies to the entire term in the parentheses, which means that:
(2^5)^4 = (2^5)*(2^5)*(2^5)*(2^5) and
(2^3)^2= (2^3)*(2^3)

This is confusing because we are dealing with two exponents. But let’s imagine that we replaced the term in the parentheses with a variable x. If we have x^4, then we know that equals x*x*x*x. This is the same operation that we have to do here. Does that makes sense? It looks like you tried to combine the exponents, which isn’t allowed in this case.

• Shristi May 21, 2018 at 6:45 pm #

hi,

lets suppose, a = (2^10) and b = (2^3),
then, according to your equation there,

(2^10)^2 + (2^3)^2 will be a^2 + b^2 and,
((2^10) x (2^3))^2 will be (ab)^2.

but a^2 + b^2 is not equal to (ab)^2.

2. nivi December 13, 2016 at 10:01 pm #

For last question, we can simplify it as 3 * 3 ^-x = 3^0 which implies adding power gives 1-x =0 which results x =1

• Magoosh Test Prep Expert December 15, 2016 at 10:36 am #

Hi Nivi,

That is correct! It is a bit more complicated than the plugging in strategy we used for this question, but it gives the same correct result 🙂

3. ana October 3, 2016 at 8:26 am #

It might be stupid from me to write this, but why cant I add until 2^26 in example 2? if they have the same base (2)

• Magoosh Test Prep Expert October 4, 2016 at 8:54 pm #

Hi Ana,

There are no such things as stupid questions 🙂 When we have like bases that are multiplied together, we can add the exponents together. If this question were 2^20*2^6 (multiplied instead of added), we would get 20^26. However, since they are added together here, we can only factor like Chris did in the explanation. We cannot combine terms with like bases using addition! Exponent rules are difficult to keep straight! This blog post might be helpful for you: https://magoosh.com/gre/2011/gre-math-basics-%E2%80%93-exponents/

4. Deepan May 12, 2016 at 6:57 am #

4^3+4^3+4^3+4^3 = 4^2(4^1+4^1+4^1+4^1)
=4^2(4+4+4+4)
=4^2(16)
Must be in this way right according to 2nd problem. Please can you explain its contradicting sorry for my ignorance.

• Magoosh Test Prep Expert May 18, 2016 at 9:41 am #

Hi Deepan 😀

Your method is a valid way to come up with the correct answer, and it turns out that your final answer is another way of writing the correct answer choice 4^4. Let’s see how 🙂

First, if we multiply out your answer of 4^2*(16) and 4^4, we will see that they are equal to the same value:

* 4^2*(16) = 256
* 4^4 = 256

And we can rewrite your answer to match the choice 3. First, let’s rewrite the quantity 16 in your answer as 4^2.

4^2*(16) = 4^2*4^2

We are left with the product of two exponents with the same base. Applying the rules of exponents, we can combine these terms by adding the exponents together:

4^2*4^2 = 4^(2+2) = 4^4

As you can see, we arrive at Choice 3, the correct answer to this problem 🙂 I hope this clears up your doubts about the practice problem! 😀

5. Julissa November 30, 2015 at 11:23 am #

I am extremely confused. for problem 3 your answer was e) 1…however when solving the formula you plugged in -1 which then forced you to turn it into a fraction. So how is the answer positive 1 if you plugged in -1?

Thank you for your time!

6. Elandra September 5, 2015 at 1:53 pm #

On problem #2, why does the outside exponent only effect the inner exponent and not the inner digit? I thought the outer exponent needed to be applied to all elements inside the parentheses. So (2^5)^4 would be (2^4)^20.

• Chris Lele September 8, 2015 at 2:44 pm #

That’s an interesting question!

First off,

(4 x 3)2 = 12(2) = 24. First, we did what was inside the parenthesis. However, were we to have first distributed the ‘2’, we would have gotten something different:

(2)(4) x (2)(3)

Notice that now we are multiplying 3×4 by 2×2, not just by 2.

When we have addition inside the parenthesis, say (4 + 3)2, then distribution works. This, however, is not the case with either multiplication or when you use exponents. Taking the original (2^5)^4, what we have is (2^5)(2^5)(2^5)(2^5). Each one of the parenthesis has five ‘2’s (I won’t write these all out :)). Therefore, there are a total of twenty 2’s. Or, in other words, I have four parenthesis (4) and five (5) 2’s in each parenthesis: 4 x 5.

I hope that makes sense 🙂

7. Varun May 24, 2015 at 7:32 am #

Hi,

Can anyone help me to understand this…

if N=2^64
N^N= 2^K …………… Find K=….?

Thanks,

• Randa August 1, 2015 at 9:16 am #

Hi Varun,

Did you manage to find an answer to this question? I am stumped as well. 🙁

8. paul nwabuona January 8, 2015 at 11:31 am #

I got all three.
im so happy cos i used to be a total mathematics noob!!!……
Now i am getting better in maths and maths problem don’t look like latin anymore.
Its not easy but i am glad i am committing time and effort into this phase of my life.
Hopefully, i will make my dream school. If not, i am eternally grateful to you guys.

9. Aaron October 20, 2013 at 12:32 pm #

Just noting a typo (I think) in your current answer explanation.

“The next problem also asks you to add exponents, but first, you must multiply the exponents in parenthesis to the exponent outside the parenthesis, giving you: 2^20+26.”

I think it’s supposed to give you 2^20 + 2^26.

10. Nitish April 6, 2013 at 6:42 am #

We can also solve the third sum as

1) 3^-x(1+1+1)=1

2) 3^x (3^1)=1

3) so 3^1-x=1

so for ans to be 1 it has to be 3^0 .

THEREFORE x=1 !!

• Chris Lele April 9, 2013 at 1:02 pm #

Hi Nitish,

I’m not sure how you arrived at your first step. 3^(-3x). This doesn’t seem to match the original equation. Maybe I’m not seeing something 🙂

• Nitish April 10, 2013 at 11:07 am #

Oh i’m sorry i have made mistake in it …i didn’t notice it 🙂

Actually i have taken 3^-x as common as follows :

3^-x (1+1+1) taking common dis way

and den we will get 3^-x (3^1)

Now as the bases are same we can add their exponents as follows

3^-x+1

so now for the L.H.S=R.H.S we need to substitute x so that its answer comes as 1

So when x=1

it becomes 3^0 which is 1

Dats d answer 🙂

• Chris Lele April 11, 2013 at 3:48 pm #

Hi Nitish,

It looks like 3^-x(3^1) is different from (3^-x)(3^1). So that way you can’t just keep the base and add the exponents. Check out RM’s solution below. Or, you can just plug in a value for x.

It’s always a good idea to start small; x=1 gets you to the answer right away. It’s fun using this method – and you are much less likely to make a mistake along the way :).

11. RM July 12, 2012 at 10:06 pm #

Hi Chris,

The 3rd question, I solved it this way, I found this easy.

3^-x + 3^-x + 3^-x = 1
3^(-x+1) = 3^0
Since, both the base are same, the exponents will be equal
-x+1 = 0
x= 1

• Chris July 13, 2012 at 4:26 pm #

Hi RM,

As they say, whatever works for you :). Being able to notice that 3^-1 equals 1/3 and you have three 1/3 or 3(1/3) though can make things go by really quickly. You don’t even have to write down a single step!

12. Krishna July 25, 2011 at 2:12 am #

In the third question, I find solving the equation easier than doing trails:
3*3^(-x)=1
3^(1-x)=1
so 1-x=0 and hence
x=1

• Chris Lele July 25, 2011 at 3:16 pm #

Solving it directly, if you are comfortable with exponents, definitely works too. There are a few extra steps though. I think noticing that each fraction equals 1/3 if x =1 is the quickest way. Again, there are multiple approaches to solving these problems. Find which way works for you is best.

13. Mohamed July 12, 2011 at 8:38 am #

1) 4^3 (4) = 4^4,

2) 2^20 + 2^6 = 2^6 (2^14+1),

3) tricky but 1 FTW

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