Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
rejoice21
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(3^5x + 3^5x + 3^5x)

by rejoice21 Sun Mar 24, 2013 9:30 pm

CAT 5/#15

(3^5x + 3^5x + 35x)(4^5x + 4^5x + 4^5x + 4^5x) = ?

A) 12^5x+1
B) 3^15x+4^20x
C) 25^5x
D) 7^35x
E) 25^5x+1

The answer explanation is as follows:
Step 1: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =
Step 2: 3^5x(1 + 1 + 1) × 4^5x(1 + 1 + 1 + 1) =
Step 3: 3(3^5x) × 4(4^5x) =
Step 4: 3^5x+1 × 4^5x+1 =
Step 5: (3 × 4)^5x+1 =
Step 6: = 12^5x+1

I don't understand what happened between Step 3 and 4. How did the exponent become ^5x+1 for both 3 and 4? What happened to 3^5x and 4^5x?
jnelson0612
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Re: (3^5x + 3^5x + 3^5x)

by jnelson0612 Sun Mar 24, 2013 10:15 pm

rejoice21 Wrote:CAT 5/#15

(3^5x + 3^5x + 35x)(4^5x + 4^5x + 4^5x + 4^5x) = ?

A) 12^5x+1
B) 3^15x+4^20x
C) 25^5x
D) 7^35x
E) 25^5x+1

The answer explanation is as follows:
Step 1: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =
Step 2: 3^5x(1 + 1 + 1) × 4^5x(1 + 1 + 1 + 1) =
Step 3: 3(3^5x) × 4(4^5x) =
Step 4: 3^5x+1 × 4^5x+1 =
Step 5: (3 × 4)^5x+1 =
Step 6: = 12^5x+1

I don't understand what happened between Step 3 and 4. How did the exponent become ^5x+1 for both 3 and 4? What happened to 3^5x and 4^5x?


Sure! Think about it this way:
What happens if I multiply (3^3) * (3^2)? That equals 3^5. How did I get that? I added the exponents and kept the base the same.

By the same logic, think about 3(3^5x). Let's think about that "3" at the start as 3^1. Thus, I really have (3^1)(3^5x). That's 3^(1+5x) = 3^(5x+1).

Using this line of reasoning, I take 4(4^5x) and convert it to (4^1)(4^5x), which becomes 4^(5x+1).

Does this make more sense now? Let us know if we can help further. :-)
Jamie Nelson
ManhattanGMAT Instructor
rejoice21
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Re: (3^5x + 3^5x + 3^5x)

by rejoice21 Sun Mar 24, 2013 10:39 pm

A case of making something simple look complicated!

Thank you, Jaime!
RonPurewal
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Re: (3^5x + 3^5x + 3^5x)

by RonPurewal Mon Mar 25, 2013 2:23 am

rejoice21 Wrote:A case of making something simple look complicated!

Thank you, Jaime!


good stuff.

this kind of thing is a big part of "harder" math problems - i.e., they use the same actual mathematics as the "easier" problems, but with more unusual-looking expressions substituted in.

if you can get to a point where you see (3)(3^(5x)), (3)(3^4), and (3^2)(3^4) as "all in the same bucket" - not as fundamentally different things - then your job there is done.