Exponent Challenge Question Series: Day 2

by on September 27th, 2010

This is the second in a series of five blog posts that we will be publishing this week highlighting challenging exponent questions!

This week on Beat The GMAT, Veritas Prep’s authors will each day contribute a difficult exponent-related challenge problem, with a solution to follow the next day. Before you begin, you may want to consider this as a way to crack the exponent code; nearly all exponent-based problems can be solved using a combination of these three guiding exponent principles:

  1. Exponent rules are almost all related to multiplication and division with virtually no rules that directly apply to addition and subtraction. When facing exponent problems, look for opportunities to factor and multiply to put yourself in a position to use your multiplication-heavy exponent expertise.
  2. Most exponent rules require you to have common bases in order to apply them, so look to break down bases into prime factors so that you have common bases with which to work.
  3. Exponents are very pattern-driven, so when large numbers are present you can often establish a rule using small numbers and then extrapolate it to the larger ones.

Yesterday’s Challenge Problem and Solution

If you haven’t don’t so already, try out yesterday’s problem first before reviewing the solution.

If 8^{x}9^{2y}=81(2^{12y}), what is the value of x?

(A) 2
(B) 4
(C) 8
(D) 12
(E) 16

B. In this problem, break down the larger numbers to find common bases for the exponents. 8^{x} can be written as (2^{3})^{x} and 81 can be written as 9^{2}, yielding (2^{3})^{x} * 9^{2y} = 9^{2}(2^{12y}). Because we now have common bases for the 9s, we can ignore (2^3)^x and 2^{12y} because 2 to any power does not share any factors with 9 to any power (since 9 = 3 x 3).  Therefore, we can determine that 2y = 2, and that y = 1.

Plugging in for the 2 terms, we then have (2^{3})^{x} = 2^{12}.

As an exponential rule, (2^{3})^{x} is equal to 2^{3x}, so we can again set the exponents with same bases equal to find that 3x = 12, and that x = 4.

Today’s Challenge Question

15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}

(A) 2^{7}
(B) 2^{8}
(C) 2^{9}
(D) 15(2^{7})
(e) 15 (2^{8})

15 comments

  • first come with denominator.
    =2^-5(1+1/2+1/4+1/8)=1/32(15/8)=15/(2^8)
    Now with the whole
    15/(15/2^8)=2^8=B

  • 2^8

    • =15*(2^5) + (2^6) + (2^7) + (4^4),
      =15*(2*2^4) + (2*2*2^4) + (2*2*2*2^4) + (2*2^4)
      =15* 2^4 * (2+4+8+2)
      =15*2^4 * (16)
      =15*2^4*(2*2*2*2) --> 15*2^4 * 2^4
      =15* 2^8

      --> E

    • (B) 2^8

    • answer B= 2^8

  • B 2^8 (15 / 15*2^-8)

  • 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}
    = 15 / {{2^{-5}+2^{-6}+2^{-7}+2^{-8}}}
    = 15 /{2^{-8} {2^3+2^2+2^1+1}}
    = 15x{2^8}/(8+4+2+1)
    = 15x {2^8} / 15
    = {2^8}
    = 256 Ans.

  • The correct answer is B.

    Since by simplifying, the denominator gives value 15/2^8.

    And in total 15/(15/2^8) gives the answer 2^8.

    Remember that 4^-4 = 2^-8.

  • The correct answer to this one is (A) 2^7.

    Solution:
    Denominator -> 2^(-5) + 2^(-6) + 2^(-7) + 2^(-4)
    = [2^(-4)] * [2^(-1) + 2^(-2) + 2^(-3) + 1]
    = [2^(-4)] * [1/2 + 1/4 + 1/8 + 1]
    = [2^(-4)] * [{4+2+1+8}/8]
    = [2^(-4)] * [{15}/8]
    = [2^(-4)] * [{15}*2^(-3)]
    = [2^(-4-3)] * [15]
    = [2^(-7)] * [15]
    Numerator -> 15
    Hence answer is:
    15/([2^(-7)] * [15])
    =1/[2^(-7)]
    =2^7 <----(A)

    • oops... didn't notice the 4 in the last term...
      the answer is 2^8.
      My bad!!!

    • don't mess up with these silly mistakes in real GMAT paper... bingo!

  • the answer is A. 2^7.

    denominator= 2^-4(1/2+1/4+1/8+1)
    so, answer= 15/2^-4(1/2+1/4+1/8+1)
    =15*16*8/(4+2+1+8)
    =15*16*8/15
    =2^7

  • one more B in the list..

  • Answer is B : 2^8

  • Using reciprocals in the denominator, the answer is 2 to the power 8.

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