Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B How do you solve this simple problem: (-2)^(n-1) = 2^8

  1. Apr 14, 2017 #1
    How do you solve this..

    (-2)^(n-1) = 2^8

    The answer is 9 but how do you get it?
  2. jcsd
  3. Apr 14, 2017 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You could write the left side as (-1 x 2)^(n-1) = -1^(n-1) x 2^(n-1). Does that help?

  4. Apr 14, 2017 #3
    OK, so continue..what is the proper solution?
  5. Apr 14, 2017 #4


    User Avatar
    Science Advisor
    Gold Member

    Solve by inspection :

    The sign must be the same on both sides so (n-1) must be even and (-2)^(n-1) is the same as ...... ?
  6. Apr 14, 2017 #5


    Staff: Mentor

    We're here to help you, but only by guiding you to reach the solution.
  7. Apr 14, 2017 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You should try to work it out. Since (-1)^8 = 1, you can multiply both sides by 1 this way:

    (-2)^(n-1) = (-1)^(8) x 2^8 = (-2)^8

    Does that help?

    Last edited: Apr 14, 2017
  8. Apr 14, 2017 #7
    This seems to work. However, I wonder if this is considered a general solution. Is there any other way to solve other than knowing that -1^8 = 1? Could I change the problem slightly where this trick (-1^8 = 1) wouldn't work?
    I originally thought about using logarithms but you can't take the log of a negative number.
  9. Apr 14, 2017 #8
    At ##\text[ \text{I}\text]## level this shouldn't be a problem.
    ##log(-2)=log(2)+\pi i+2 k \pi i## for ##k\in Z##.
  10. Apr 14, 2017 #9
    I think perhaps this is a trick question, in the sense that the answer is obvious, if you think about it before just mechanically grinding out the algebra.

    The first thing to note is that -2^m = 2^m if m is even.

    -2^2 = -2 * -2 = 4
    -2^3 = 4 * -2 = -8
    -2^4 = -8 * -2 = 16
    -2^5 = 16 * -2 = -32
    -2^6 = -32 * -2 = 64
    -2^7 = 64 * -2 = -128
    -2^8 = -128 * -2 = 256 = 2 ^8

    This should be clear if you know that multiplying two negative numbers results in a positive number.

    Of course you should not need to write the above equations. Really you are just solving for n-1 = 8, therefore n = 9.

    In general, if you take the standardized math tests like the SAT or GRE, or even tests in school, sometimes a little thought saves a lot of grind work.

    [Edited to correct some typos, which can happen when one is in a hurry to write down the obvious!]
    Last edited: Apr 14, 2017
  11. Apr 14, 2017 #10


    Staff: Mentor

    No, this isn't true. The left side is a negative number, and the right side is a positive number.

    The usual order of operations is that exponents are evaluated before the negation, so -2^2, say is -4, while (-2)^2 is 4.
  12. Apr 14, 2017 #11


    Staff: Mentor

    I believe that this is really a B-level thread, not I-level, so I have changed the level to "B".
  13. Apr 14, 2017 #12
    You mean -2^2 = -(2^2) = -4, or in general -2^(n-1) = -(2^(n-1)).

    But he wrote (-2)^(n-1), not -(2^(n-1)).

    So applying PEMDAS I interpret this to mean he is applying the exponent n-1 to -2.
  14. Apr 14, 2017 #13


    Staff: Mentor

    My comment was relevant to what you wrote; namely
    and I included your quote in my post (#10).
    -2^m is a negative number while 2^m is positive.
  15. Apr 14, 2017 #14
    I think you are right. I'm used to doing this as I did, since I have thought it's clear from the context. But that does not make it right.

    I did some quick "research" to see what Dr. Math says. He would agree with you.


    Thanks for pointing this out. I will correct my answer to the following.

    The first thing to note is that (-2)^m = 2^m if m is even.

    (-2)^2 = (-2) *( -2) = 4
    (-2)^3 = 4 *(-2) = -8
    (-2)^4 = (-8) * (-2) = 16
    (-2)^5 = 16 * (-2) = -32
    (-2)^6 = (-32) * (-2) = 64
    (-2)^7 = 64 * (-2) = -128
    (-2)^8 = (-128) * (-2) = 256 = 2 ^8

    Thanks for the correction.

    From Dr. Math:

    "When you have numbers only, as in -9^2, it's not at all clear that we
    should treat it differently from -x^2. However, some will argue that
    it should, because -9 represents a single number, not an operation on
    a number. Thus, some will interpret -9^2 as (-9)^2, while others will
    read it as -(9^2)."

    So I was in the "some will argue" camp. But if there is any possibility of confusion, we should use parentheses. So I am going to accept your correction.

    Now I will be on the lookout for this tendency everywhere. :)
  16. Apr 14, 2017 #15
    p.s. now it would be interesting to see if, in mathematical papers or textbooks, there is a tendency to distinguish between -x^n and -2^n, in the interpretation sense mentioned by Dr. Math. I suspect the reason I use the convention I do is because I learned it by example, either from books or in the classroom. Very interesting.
  17. Apr 14, 2017 #16


    Staff: Mentor

    The post from Dr. Math that you cited made me think, as well. I've known for many years that -x^2 means the negative of the square of x. I assumed that the same idea applied for constants, as in -2^4. It didn't occur to me that some would distinguish between the negation operation on 2 vs. -2 as a number in its own right.

    I looked for some web sites to see if there are varying treatments of, say -2^2. Here's one I found that has a calculator (https://www.symbolab.com/solver/simplify-calculator/simplify -2^{2}), and that treats -2^2 as if it were written -(2^2).
    The explanation shown under Steps is horrible.
    ##2^2 = 4##
    ##= -4##
  18. Apr 19, 2017 #17
    Reason I noticed this thread is that my current project as an amateur is to review my high school algebra (via a battered copy of Algebra: Structure and Method, by Brown and Dolciani); and in working my way through the first few chapters, I've realized something I probably neither noticed nor cared about 45 years ago - namely how easy it is to get bitten by notation involving negative signs, as has been discussed here. Another interesting example of this (at least to me, 45 years later) is that a negative sign in front of a variable needn't mean the variable itself is negative.

    I am curious though about one thing. I agree w/ @Aufbauwerk 2045 about this problem seeming almost deliberately trivial; so @barryj, can I ask where you found it and what the context was?
    Last edited: Apr 19, 2017
  19. Apr 20, 2017 #18
    In reply to the original post: ( I don't know if this has already been mentioned) as the bases are the same on both sides you can equate the exponents. This gives n-1=8 and hence n=9.
  20. Apr 20, 2017 #19


    Staff: Mentor

    The bases aren't the same on both sides. On the left the base is -2, and on the right it's 2.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: How do you solve this simple problem: (-2)^(n-1) = 2^8