9^3(2^8 + 2^9/2) =

\(9^3*\frac{2^8+2^9}{2}\)
\((3^2)^3*\frac{2^8*(1+2)}{2}\)
\(3^6*\frac{(3*2^8)}{2}\)
\(3^6*3*2^7\)
\(3^7*2^7\)
\(6^7\)

\(9^{3}(\frac{2^{8} + 2^{9}}{2})\)

= \(3^{6}2^8(\frac{1+2}{2})\)

= \(3^{6}2^8(\frac{3}{2})\)

= \(3^7 2^7\)

= \(6^7\)

Answer:- C

VERITAS PREP OFFICIAL SOLUTION:

When working with exponents, you should try to find common (typically prime) bases and multiply. First to multiply, you can factor a 2^8 out of each additive term within the parentheses to get: \(9^{3} 2^{8} (\frac{(1+2)}{2})\). Then, perform the addition within the parentheses to get: \(9^{3}2^{8} (\frac{3}{2})\).

You can break down \(9^{3}\) to be \((3^{2})^3\), which is \(3^{6}\), and then you have common prime terms for each exponent: \(3^{6}2^{8} (\frac{3}{2})\).

Multiplying out the equation \(3^{6}2^{8} (\frac{3}{2})\), you get: \(3^7*3=3^8\) and \(\frac{2^{8}}{2} =2^7\), which can be simplified to \(3^7*2^7\).

This can be expressed as \((3*2)^{7} = 6^{7}\), and the correct answer is C.

\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

\(= 9^{3}(2^{7} + 2^{8})\)

\(= 9^{3} * 2^7 (2 + 1)\)

\(= 9^{3} * 2^7 * 3\)

\(= 3^{6} * 2^7 * 3\)

\(= 3^{7} * 2^7\)

\(= 6^{7}\)

Hence, Answer is C

We can re-express 9^3 as (3^2)^3, or 3^6, and we can re-express the fraction (2^8 + 2^9)/2 as two separate fractions. Thus, we have:

3^6[(2^8/2) + (2^9/2)]

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

3^6[2^7(1 + 2)]

3^6[2^7(3)]

3^7 x 2^7 = 6^7

Answer: C

\(3^{6}(\frac{2^{8}(1 + 2)}{2})\)

Or, \(3^{6}*2^{7}*3\)

Or, \(6^7\), ANswer must be (C)

9^3*((2^8+2^9)/2)=
3^6*(2^7+2^8)=
3^6*2^7*(1+2)=
3^6*2^7*3=
3^7*2^7=
6^7

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.