# Exponent Challenge Question Series: Day 3

This is the third 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:

- 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.
- 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.
- 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 Question and Solution **

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

(A)

(B)

(C)

(D)

(e)

**B.** When you face exponents and addition, you’ll almost always need to factor out common exponential terms so that you can multiply, which is the operation that lends itself to more flexibility with exponent rules. Here, you can first take and express it as , which equals , so that you have all common bases in the denominator.

Then, in order to multiply, factor out a from each term to get: . The is now a multiplicative term, which you can express also as , allowing you to flip it to the numerator, which becomes .

Back to the denominator, you can express as which equals .

Putting the entire fraction together, the fraction is . The 8 can be expressed as , again providing a common exponential base, making the fraction / . Dividing by a fraction is the same as multiplying by its reciprocal, leaving . The 15s cancel, leaving which equals , answer choice B.

**Today’s Challenge Question**

(A)

(B)

(C)

(D)

(E)

## 11 comments

ikaplan on September 28th, 2010 at 6:08 am

9^3[2^8+2^9/2]=

3^6[2^8(1+2)/2]=

3^6*2^8*3*2^-1=

3^6*3*2^8*2^-1=

3^7*2^7=

(3*2)^7=

6^7

Hence the answer is C

saurabh goyal on September 28th, 2010 at 6:08 am

9^3= (3^2)^3 = 3^6

2^8(1+2)/2 = 2^8 (3)/2 = 2^7 * 3^1

in all 3^6.2^7.3^1 = 3^7.2^7 = 6^7

option C

Yogesh on September 28th, 2010 at 6:24 am

Ans is C: 6^7

3^6(2^7 + 2^8)

=3^6 * 2^7(1 + 2)

=3^7 * 2^7

=6^7

Kerby Altidor on September 28th, 2010 at 6:59 am

(C) 6^7

shibbirahamed on September 28th, 2010 at 7:45 am

i agree u.

ASKI@1st on September 28th, 2010 at 7:54 am

(3^2)^3 *(2^7+2^8)=3^6*2^7*(1+2)

= 3^6*2^7*3

= 3^7*2^7 = 6^7 (C)

Saptarshi Roy on September 28th, 2010 at 10:03 am

yep its 6 ^ 7.

Anahatha on September 28th, 2010 at 10:37 am

C) 6^7

saurabh goyal on September 28th, 2010 at 11:06 am

option C.

6^7

Ishaan Singhal on September 28th, 2010 at 11:18 am

Answer (C) 6^7

Solution:

9^3 * ([2^8 + 2^9]/2)

=[(3^2)^3] * ({2^8 * [1 + 2]}/2).......Taking 2^8 common and 9=3^2

=[3^(2*3)] * (2^7 * 3)

=[3^(6)] * (2^7 * 3)

=3^7 * 2^7

=(3 * 2)^7.........since the exponents are same we can multiply the bases

=6^7.............(C)

al on November 15th, 2011 at 2:58 pm

i agree; tough to get it within 3 minutes