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.

2.

3. If

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,

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:

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

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!

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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

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 🙂

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)

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: http://magoosh.com/gre/2011/gre-math-basics-%E2%80%93-exponents/

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.

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*4^2We 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! 😀

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!

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.

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 🙂

Hi,

Can anyone help me to understand this…

if N=2^64

N^N= 2^K …………… Find K=….?

Thanks,

Hi Varun,

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

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.

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.

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 !!

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 🙂

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 🙂

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 :).

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

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!

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

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.

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

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

3) tricky but 1 FTW