Adding and Subtracting Powers on the GMAT

Adding and Subtracting Powers GMAT-magoosh

First, consider this practice problem:

  1. \(7^6 + 7^8\)

If this problem makes your head spin, you have found the right blog article!

Operations with powers

There are legitimate laws of exponents for multiplying and dividing powers.

Take a look at these examples:

  • \(a\times(b + c) = a\times b + a\times c\)

  • \((a^m) \times (a^n) = a^{m + n}\)

  • \( \frac{a^m}{a^n} = a^{m – n}\)

  • \((7^8) \times (7^6) = 7^{8+6} = 7^{14}\)

  • \( \frac{7^8}{7^6} = 7^{8 – 6} = 7^2 = 49\)

Multiplying powers means add the exponents and dividing powers means subtract the exponents. For more on this, see this post on exponent properties.

Alas! There is no law of exponents for adding and subtracting powers. 😢

In other words, \(a^n\) plus or minus \(a^m\) is not going to equal a-to-the-power-of-anything-in-particular.

There is no convenient way to combine a sum or difference of powers into a single power expression.

Distributing and factoring out

But, all is not lost! We can avail ourselves of one of the fundamental number properties, the Distributive Law, and its inverse process, “factoring out.”

That is a rather remarkable and often underestimated equation.

First of all, it is a statement of the Distributive Law. Remember \(a\times(b + c) = a\times b + a\times c\)?

If we start on the left and go to the right, we are distributing “a.” The reverse process is called factoring out: if we start on the right side, and go to the left, we are factoring “a” out.

Multiplication distributes over addition and subtraction: we could replace both plus signs with minus signs in that equation and it would still be true.

Written in simple variable form, it probably doesn’t look all that earth-shattering, but the implications of that equation are so widespread as to boggle the mind. Addition and subtraction of variables, addition and subtraction of fractions, and all techniques for factoring quadratics are ultimately dependent on this profound fundamental law.

Before discussing the implications for addition and subtraction of powers, I will just make one more point. In this above equation, “a” can be any common factor of the two terms on the right, but in some sense, we “maximize” the effect of factoring out when the factor we factor out is the Greatest Common Factor (the GCF). See this link for a review of how GCF works with numbers.

Factoring out to add & subtract powers

Suppose we want to simplify \(2^{17} – 2^{13} = ?\)

Well, since both terms are powers of two, they share several common factors (\(2^1\), \(2^2\), \(2^3\), etc.).

The GCF of both of these terms is \(2^{13}\) itself. We can express each term as the product of the GCF and something else.

\(2^{17} = (2^4) \times (2^{13})\)


\(2^{13} = (1) \times (2^{13})\)

Now, we can write the difference of powers as

\(2^{17} – 2^{13} = (2^4)\times (2^{13}) – (1)\times (2^{13}) = ?\)

I realize that, at the moment, it may not be completely obvious that this will ultimately be a step in the direction of simplification — it looks like we made everything gratuitously more complex in the second step! Trust that things will simplify soon! Here’s the Distributive Law equation again, now with subtraction and with left & right switched from above.

\(a \times b – a \times c = a \times (b – c)\)

Notice that the second step of the power subtraction equation matches perfectly the pattern on the left of this statement of the Distributive Law: \(a = 2^{13}\), \(b = 2^4\) and \(c = 1\). Following this pattern, we see this turns into:

\(2^{17} – 2^{13} = (2^4)\times (2^{13}) – (1)\times (2^{13}) = (2^{13})\times [(2^4) – 1]\)

Now, it’s germane to point out: we should be able to simplify \(2^4\).

\(2^1 = 1\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)

That, in turn, allows us to simplify the piece in the brackets in the last expression:

\(2^{17} – 2^{13} = (2^4)\times (2^{13}) – (1)\times (2^{13}) = (2^{13})\times [(2^4) – 1]\)
\(= (2^{13})\times [16 – 1] = (2^{13})\times (15) \)

Voila! That final expression, fifteen times two to the thirteen, is the simplification of the difference of two powers.

This process would be particularly important if, for example, you were asked to simplify something like:

\(\frac{[2^{17} – 2^{13}]}{20}\)

We can’t cancel from either term in the numerator as long as we have addition or subtraction in the numerator, but once we simplify the numerator so that it’s all multiplication, we can cancel. You should verify for yourself that:

\(\frac{[2^{17} – 2^{13}]}{20} = (3)\times (2^{11})\)

A remarkable simplification!

Further Practice

Now that you have seen that, take another look at the practice question from above, before reading the answer and explanation below.

Practice question #1

Remember the first practice question you encountered before this article began?

  1. \(7^6 + 7^8\)

Click here for the answer and explanation to practice question #1

The GCF of \(7^6 \)and \(7^8\) is \(7^6\).

\(7^6 = (1)\times (7^6)\)
\(7^8 = (7^2)\times (7^6)\)


\(7^6 + 7^8 = (1)\times (7^6) + (7^2)\times (7^6) = [1 + 7^2]\times (7^6) \)
\(= [1 + 49]\times (7^6)\) 
\(= (50)\times (7^6)\)

Answer = E

Did your answer change after you read this post? Let us know in the comments below.

Practice question #2

  1. if \(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 4^{24}\), then n

Click here for the answer and explanation to practice question #2

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  • Mike MᶜGarry

    Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as "member of the month" for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's Youtube video explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test.

21 Responses to Adding and Subtracting Powers on the GMAT

  1. Tina baid July 18, 2018 at 5:19 am #

    4power m – 4power m-1=24.find m. My answer came 2.5

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert July 25, 2018 at 7:57 am #

      Assuming you mean (4^m) – [4^(m-1)], then yes, 2.5 would be correct. 🙂 This process, of course, involves non-integer powers, a complicated operation that isn’t really covered in this post. Quora has a good thread on these kinds of exponents here.

      • Sanzid January 7, 2021 at 11:51 am #

        pls solve it

        you can write A to the power B as A^B and A divided by B as A/B, and A multiplied by B as A*B, etc

  2. Arjun June 15, 2018 at 1:59 am #

    Can you solve this

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert June 27, 2018 at 1:27 pm #

      Yes, I can solve it. 2*10^6 would be 2,000,000, and 1.2*10^-3 is 1.2*(10/10/10), or 1.2*0.001, or. From there, it’s a matter of simple subtraction. Assuming of course, that you meant the order of operations to look as follows: (2×10^6)-(1.2×10^-3). Otherwise, without the parenthesis, your default order of operations would lead to a more complicated equation and a different answer. 😉

  3. Hitesh Ghagave December 10, 2017 at 6:49 pm #

    Can you solve (T^-1 + T^2 +T^1)

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert December 21, 2017 at 1:35 pm #

      Hi Hitesh,

      Can you provide some more detail about where you found this question and the context of the question? We are not able to provide support for outside materials, but we are more than happy to help with questions related to Magoosh and Official materials. For this question, it’s important to realize that you will be dealing with a polynomial expression that will have multiple answers. You will need to solve this using a quadratic equation. As we say in this blog post, there is no ‘easy’ method for solving questions that involve addition and subtraction of powers.

  4. KAJOLSHARMA November 7, 2017 at 1:00 am #

    Pls solve it
    (1/6)^2 +(1/6)^4 =6^2(m+1)
    Than find m

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert November 7, 2017 at 10:00 am #

      An interesting equation, Kajolsharma. 🙂 A couple of questions before we look at a solution, though. First, where did you see this equation? Second, is (m+1) supposed to be part of the exponential notation, or is it supposed to be a separate expression, multiplied by (6^2). In other words, does the equation mean this:

      (1/6)^2 +(1/6)^4 =6^[2(m+1)]

      or this:

      (1/6)^2 +(1/6)^4 =(6^2)(m+1)

      With your current notation, things are a little bit ambiguous, so let me know. 😉

  5. jim coleman August 30, 2016 at 9:22 am #

    BTW what I did was to substitute 10 for 2 to start.

    and the verbiage above SHOULD have been more than, less than or equal to.


  6. Sudhan March 23, 2016 at 1:30 am #

    How do you do:


    I couldn’t work it out, please answer so I can understand a bit better!

    Cheers, Sudhan

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert March 23, 2016 at 11:25 am #

      Hi Sudhan,

      We can find the sum of 4^3+4^3+4^3+4^3 in a couple of ways. For example, recognizing that we have the same term, 4^3, repeated 4 times, we can rewrite the expression as 4*(4^3). From there, we will want to evaluate 4^3 = 64 and find the product

      4*64 = 256.

      We could also first evaluate 4^3 = 64 and add replace each term of 4^3 with 64 to find the sum:

      4^3+4^3+4^3+4^3 = 64+64+64+64 = 256.

      I hope this helps 🙂

    • Jayesh May 4, 2016 at 8:14 am #

      I am unable to solve below.
      2^x+y * 3^x-y = 32/3

  7. Neka January 29, 2014 at 2:12 pm #

    Im finding it difficult to get from (2^17-2^13/20) to the answer of (3)*(2^11). I might be doing something wrong in my process but cant figure out how you simplified to 3 and why the change from 2^13 to 2^11.

    • Mike MᶜGarry
      Mike January 29, 2014 at 2:22 pm #

      In the post above, I showed how 2^17-2^13 simplifies to (15)*(2^13). I’ll assume you follow that, because the steps are printed above. The denominator factors into
      20 = 5*4 = 5*(2^2), When we divide (15)*(2^13) by 5*(2^2), the 15/5 =3, and the (2^13)/(2^2) = 2^11.
      Does all this make sense?
      Mike 🙂

    • Neka January 30, 2014 at 7:46 am #

      Awesome. Thanks. Didn’t realize I could break down the denominator.

      • Mike MᶜGarry
        Mike January 30, 2014 at 10:24 am #

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

  8. Kate August 28, 2013 at 12:47 pm #

    That was so helpful, and not in my Kaplan book at all. Thank you!

    • Mike MᶜGarry
      Mike August 28, 2013 at 1:00 pm #

      Dear Kate,
      I’m very glad you found this helpful. Best of luck to you!
      Mike 🙂

  9. Domenico September 11, 2012 at 5:32 pm #

    7^6 (1 + 7^2) —-> 7^6 ( 1+ 49) —-> 7^6 * 50

    10 seconds 😉

    Thanks Mike

    • Mike MᶜGarry
      Mike September 12, 2012 at 10:55 am #

      You are quite welcome.
      Mike 🙂

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