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Practice Problems on Powers and Roots

Here are ten reasonably challenging problems on powers and roots.  Solutions follow the problems.   Remember, no calculator!

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Statement #1: x < 1

Statement #2: x > –1

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  1. 4
  2. 8
  3. 14
  4. 28
  5. 42

5) If A is an integer, what is the value of A?

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  1. 11
  2. 30
  3. 45
  4. 75
  5. 225

 

Powers & roots

This is a tricky set of topics.  Here are some previous blogs I have written about these ideas:

1) Laws of Exponents

2) Patterns of powers for different kinds of numbers

3) Adding & Subtracting Powers

4) Roots

5) Dividing by a Square Root

6) Mistake = Distributing a Root

If you read one of those articles and have some insights, you may want to give the problems above another glance.  If you would like to clarify anything said here or in the solutions, please let us know the comments section.

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Practice problem explanations

1) For positive numbers less than one, when we raise them to positive integer powers (square, cube, etc.), then get smaller, further from one.  When we take roots of them (square root, cube root), they get bigger, closer to one.  The higher the root, the closer it is to one.  Thus, for any x in the 0 < x < 1 range,

ppopar_img12

Now, for negative numbers, the mirror image happens.  Of course, we can’t take even roots of negative numbers (square root, fourth root, etc.) but we can take odd roots.  For numbers in the –1 < x < 0 range, negative numbers with an absolute value less than one, when we take a root, the value gets closer to –1, and the higher the root, the closer it is to –1.  Thus, for these numbers

ppopar_img13

Thus, the rule works in opposite directions in these two regions, (0 < x < 1) vs. (–1 < x < 0).  Both of these regions are included, even with combined statements, so we can give no definitive answer to the prompt question.

Answer = (E)

 

2) Statement #1: Given this statement, we could square 23,100, and that would be some number between 4 & 6 times some power of ten.  We don’t actually have to perform that calculation.  It’s enough to note that, from this statement, we can determine the value of both A and m.  Unfortunately, n remains unknown, and without n, we can’t determine the value of T.  This statement, alone and by itself, is insufficient.

Statement #2: This statement tells us nothing about A, and it doesn’t allow us to determine unique values for either n or m.  We can’t determine anything with just this.  This statement, alone and by itself, is insufficient.

Combined statements: The first statement allows us to determine unique values for A & m.  Once we know m, we can use the second statement to find n.  Once we know both A & n, we can determine the value of T.  Combined, the statements are sufficient.  Notice, we can arrive at this conclusion without a single calculation: that’s the ideal of GMAT Data Sufficiency!!

Answer = (C)

 

3) First, we need to know the formula for the square of a binomial:

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Thus:

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Now, we need to recognize that 5 to the x and 5 to the negative x are reciprocals, so their product is one:

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Also, by the laws of exponents:

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Similarly,

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Thus,

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Subtract two from both sides, and we have

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Answer = (D)

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Thus, a & b are in a ratio of 4-to-3.  Thus, we could say that a = 4n and b = 3n, for some n.  Thus, a + b = 4n + 3n = 7n = 56, so n = 8.  Well, a – b = 4n – 3n = n = 8

Answer = (B)

ppopar_img22

Statement #1 tells us that, for some unknown positive integer, N is larger than 1,000,000 and smaller than 10,000,000.  Could there be more than one power of 23 in that region?  No, because even if N were equal to the lowest possible value in the range, 1,000,000, if we multiply another factor of 23, it would bring us up to 23,000,000, which is above the high end of the range.  The ratio of the top end to the bottom end is just 10:1 = 10, which is smaller than 23, so more than one power of 23 cannot possibly fit in this range.

Thus, there can only be one integer value A that puts N in this range.  We don’t have to calculate it: it’s enough to know that the information given unique determines the value of A.  Thus, this statement, alone and by itself, is sufficient.

Statement #2 is a tautology.  A tautology is a statement that is true by definition, and thus gives no information.  The verbal statement “My book is a book” is a tautology: it tells us absolutely nothing useful about that book.  The mathematical statement x + 5 = x + 5 is a tautology: that gives us absolute no mathematical information that would allow us to solve for x; in other words, it is true for every single number on the number line.

This statement is a more sophisticated tautology.  To understand this, we need to know how to add and subtract powers.

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This statement is true for all numbers on the number line, so it contains absolutely no useful information for determining the value of A.  This statement, alone and by itself, is insufficient.

Answer = (A)

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Cube both sides.  The cube or fifth power of a negative remains negative, but the fourth power of a negative is positive.

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Because we know that x cannot be zero, we can divide by x cubed. Notice that (positive) x-to-the-fourth divided by (negative) x-cubed equals (negative) x.

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There’s no direct way to go from this to the product xy.  We have to solve for y in terms of x.

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Now, we can simply take the fifth root of both sides.

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Now, we are in a position to find the product.

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Answer = (C)

 

7) For this one, we need to know how to simplify square roots.  We can use the formula

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Suppose we cleverly factor the number under the square root so that Q is a perfect square: then we could simplify that factor.

Notice that all three numbers under the radical sign are of the form: two times a perfect square.  That gives us a big clue to simplifying them.

ppopar_img31

Answer = (A)

 

8) Start with the given equation

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Cube both sides.

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Now, square both sides.

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Now, one solution is x = 0, but that’s not listed as an answer.  We are looking for a non-zero answer, so if x is not zero, we can divide by x.

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Now, take the fifth-root of both sides.

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Answer = (D)

 

9) With this one, it’s helpful to rewrite the roots as a fractional exponents:

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Because x ≠ 0, we can divide all three terms by ppopar_img45.  This leaves:

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This is now an ordinary quadratic.  To see this, use the substitution

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(u + 7)(u – 3) = 0

u + 7 = 0      OR     u – 3 = 0

u = –7          OR     u = +3

We have to be careful here.  Here, –7 is a solution for u, but not for x.  Now that we have values for u, we have to solve for the corresponding values of x.

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The first is not an available answer choice, but the second is answer choice (E).

 

10) Start with the first equation

ppopar_img41

Square both sides:

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Now, multiply this by the second equation:

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Answer = (C)

 

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20 Responses to Practice Problems on Powers and Roots

  1. Sebastian October 27, 2017 at 2:19 am #

    Hi Mike,

    Regarding Q3: How did you come up with the approach to solve that question with square of a binomial? It seemed counterintuitive to me to square everything.

    Thanks
    Sebastian

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert October 29, 2017 at 12:07 pm #

      Hi Sebastian,

      The best way to deal with tricky questions like this is to develop number sense, which is your ability to see potential in these sorts of equations and be able to sketch a path forward. Number sense is different for everyone, and you will develop your own personal sense of how things should go. For me, the key here was to realize that 25=5^2, which means that we will have to square the numbers in order to get to our desired equation. Once we recognize this, even if we don’t have a complete view of the path forward, we can square both sides of the equation and see what we get. This means using the square of a binomial equation. Once we square both sides, it just becomes a matter of manipulating the equation to get our desired result!

      I hope this helps–I encourage you to play around with this question and see if you can get a better sense of how the numbers work together. This will help you to understand these concepts better and allow you to really develop your number sense skills!

  2. stephenie April 5, 2016 at 8:21 am #

    Hey Mike,

    I need help with the 9th question, how did we divide all three terms by 1/6 and resulted in x 2/3+4x 1/3-21=0.

    Thanks

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert April 17, 2016 at 6:15 am #

      Hi Stephenie,

      Happy to help 🙂

      The first step in this question is to rewrite the terms as fractional exponents:

      x^(5/6) + 4*x^(1/2) – 21*x^(1/6)

      At this point, it is useful to write all of the fractional exponents using the common denominator 6:

      x^(5/6) + 4*x^(3/6) – 21*x^(1/6)

      We’re allowed to divide the terms by x^(1/6) because x is not equal to zero (knowing this guarantees that we won’t be dividing by zero, which is an illegal move). Let’s carry out the operation separately for each term. And remember, when we divide exponential terms with the same base, we subtract the exponential part:

      x^(5/6) / x^(1/6) = x^(5/6 – 1/6) = x^(4/6) = x^(2/3)

      4x^(3/6) / x^(1/6) = 4x^(3/6 – 1/6) = 4x^(2/6) = 4x^(1/3)

      -21x^(1/6) / x^(1/6) = -21x^(1/6 – 1/6) = -21x^(0) = -21*1 = -21

      Putting these terms back together, we have

      x^(2/3) + 4x^(1/3) -21

      and can now use u substitution, as Mike explains in the post, to solve for x.

      Hope this helps! 🙂

  3. Sowkhya August 23, 2015 at 11:21 am #

    Hi Mike. Great post, as always 🙂 I needed more help with the 5th question – Statement #1. The question says A is an integer. If I were to assume A as -1 instead of a positive integer 1, how and if, could I proceed with my understanding and solution to the problem?
    Thanks,

  4. Arti June 7, 2015 at 11:59 am #

    Hi Mike, Can you please help me better understand Question 2?

    How can statement 1 help us determine the value of A if A can be anything between 4 and 6? Wouldn’t we have to know the exact value of A in order to get the exact value of T?

    Thanks,
    Arti

    • Mike MᶜGarry
      Mike McGarry June 8, 2015 at 4:11 pm #

      Dear Arti:
      I’m happy to respond. 🙂 Think about what statement #1 in problem #2 says:
      sqrt(A x 10^m) = 23,100
      When we square both sides, we get:
      A x 10^m = (23,100)^2
      What you need to appreciate is that (23100)^2 is a very specific number. We aren’t able to calculate it easily without a calculator, but nonetheless, mathematically it is a very specific and unambiguous number. When we write that number in scientific notation, we would take out all the whole number powers of ten—that would be the 10^m part. The part that is left after factoring out the powers of ten would be A, and that would be a very specific number. Again, without benefit of a calculator or a computer, we ourselves cannot conveniently compute that value, but that is NOT what GMAT DS is about. It’s not about the limits of our calculation abilities without a calculator. Instead, it is about what is mathematically determined, what is determined, as it were, in the mind of God. In the mind of God, when we square 23,100, we get some definite number, and when we remove all the powers of ten, what’s left is the unique value of A. In other words, A is completely determined, even if you and I don’t know what that value is without a calculator.
      The philosopher Arthur Schopenhauer (1788-1860) said, “Every person takes the limits of his or her own field of vision for the limits of the world.” That is one of the biggest traps on the GMAT DS. If something is definite and determined, even if we can’t conveniently calculate it, then it is still sufficient. It has absolutely nothing to do with whether you could perform the calculation yourself with nothing more than pencil and paper. It has everything to do with whether it is mathematically determined, whether with sufficient resources, the calculation would be possible.
      Does all this make sense?
      Mike 🙂

      • Neil January 18, 2016 at 9:40 am #

        Hi Mike,

        Thanks for all of your posts and your responses to questions. This blog has been of immense help in my GMAT prep.

        Your reasoning for Q2, while sound, is very abstract and was difficult for me to grasp. Another approach could be:

        Statement 1:
        – square both sides and substitute A with (T/10^n) as taken from question stem
        – (T/10^n)*10^m = 23100^2
        – T*10^(m-n) = 23100^2
        – T=(23100^2) / 10^(m-n) Insufficient

        Statement 2:
        – multiply both sides with -1 to get m-n = -7 insufficient

        Substitute result of statement 2 into statement 1 to attain sufficiency. Therefore C is the correct ans.

  5. Marina October 30, 2014 at 11:13 pm #

    Hello Mike,

    In question 1, in Statement 1, how do I know that x<1 does not include all numbers smaller than 1? For example, numbers like -1, -2. In your explanation you are limiting the x<1 to the range 0 < x < 1, however it is not stated in the statement.

    • Mike MᶜGarry
      Mike October 31, 2014 at 10:21 am #

      Dear Marina,
      Great question, and I am happy to respond! 🙂 In the solution to #1, I was not being so methodical — first Statement #1 by itself, then Statement #2 by itself, and then Combined. I could tell right away that it would be unlikely that anything would be sufficient, so I jumped immediately to the two-statement scenario and showed why there was no sufficiency even with all information. Yes, theoretically, larger negative numbers would be possible with Statement #1, but if we can show that numbers in 0 < x < 1 and in -1 < x < 0 give opposite answers, it's superfluous to consider any other numbers. On GMAT DS, you don't necessarily have to explore every possibility — as soon as we find a contradiction, we are done.
      Does this make sense?
      Mike 🙂

  6. Josh July 22, 2014 at 4:01 am #

    Mike,
    In question 3 – the question has 5^x + 5^-x = B

    Why do you square the binomial?
    Using distributive law with the question as written could you not multiply both sides by 5 and arrive at 25^x + 25^-x ? I’m guessing I’ve misapplied the distributive rule here…
    Thanks

    • Mike MᶜGarry
      Mike July 22, 2014 at 10:38 am #

      Dear Josh,
      I’m happy to respond. 🙂 That question is loaded with trap answers, and I am sorry to say: you fell for one of the traps. It’s very important to appreciate the nature of the trap and understand it inside-out, so you don’t make this same mistake on the GMAT.
      Think about just 5*(5^x), just that without any addition or distribution. Think about what (5^x) is — whatever the number x is, 5^x is that many factors of 5, all multiplied together. For the sake of argument, let’s pretend that x = 300, just to pick a number. Then 5^x is 300 factors of 5, all multiplied together. What happens when we multiply this by 5? Well if we were allowed to write the product as 25^x, think about what that would mean. We would now have 300 factors of 25, and each factor of 25 has two factors of 5, so that would be a total of 600 factors of 5!! Do you see the problem? We started with 300 factors of 5, multiplied by one more factor of five, and wound up with 600 factors of five. That’s bad accounting! We can’t multiply one number with an exponent times another number without an exponent.
      Instead, 5 = 5^1, and (5^1)*(5^x) = 5^(x + 1), which properly denotes that, when we multiply by only one factor of five, the total number of factors of 5 correctly goes up by just one.
      To square that expression, we would need to use the algebra formula for the square of a binomial. See:
      http://magoosh.com/gmat/2013/three-algebra-formulas-essential-for-the-gmat/
      Remember that (5^x)^2 = 5^(2x) = (5^2)^x = 25^x. When we square the expression 5^x, we in fact do double the number of factors of five we have. This legitimately doubles the number of factors, which can be expressed as 25^x.
      Does all this make sense?
      Mike 🙂

      • Josh July 22, 2014 at 11:04 am #

        Makes perfect sense! Thanks.
        Many alarm bells were ringing when I picked the “cop-out” answer and ignored a pretty obvious 2/3 split on squares…

        The explanation is great. Thanks for your help and all of these posts they have been immensely useful.

        • Mike MᶜGarry
          Mike July 22, 2014 at 11:34 am #

          Dear Josh,
          You are more than welcome, my friend. I am very glad you have found these posts helpful. Best of luck to you!
          Mike 🙂

  7. Sriram June 25, 2014 at 8:18 am #

    Hi Mike,
    In the solution for problem 6, you say – “There’s no direct way to go from this to the product xy. We have to solve for y in terms of x ” . Can you please clarify if this is only because the answer choices for the product xy are in terms of x?

    Thanks,
    Sriram

    • Mike MᶜGarry
      Mike June 25, 2014 at 1:27 pm #

      Dear Sriram,
      Exactly! Of course we could solve x(y^5) = 1 to get xy = 1/(y^4), but that would not help us at all to select one of the answers, precisely because all of the answers are in terms of x. Does this make sense?
      Mike 🙂

      • Sriram June 25, 2014 at 2:31 pm #

        Yes it does. Thanks Mike.

        • Mike MᶜGarry
          Mike June 25, 2014 at 2:54 pm #

          Dear Sriram,
          You are quite welcome, my friend. 🙂 Best of luck to you!
          Mike 🙂

      • Neha_K November 17, 2014 at 10:34 pm #

        Hi Mike,
        For question 6, is solving this way correct —-

        Cubing both sides we get ==> x^4.y^5 = x^3
        Then multiplying both sides by x ==> x^5.y^5 = x^4
        which can be written as (x*y)^5 =x^4
        Taking fifth root on both sides, x*y = x^4/5

        Thank you for all your help!
        Neha

        • Mike MᶜGarry
          Mike November 18, 2014 at 10:16 am #

          Dear Neha,
          My friend, not only is that completely correct, but it’s actually considerably more elegant than the solution I provided. 🙂 Excellent work! I am very glad you found this blog helpful! Best of luck to you!
          Mike 🙂


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