BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Good exponent problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

Good exponent problem

by cramya » Thu Mar 12, 2009 8:31 pm
( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

Re: Good exponent problem

by x2suresh » Thu Mar 12, 2009 9:26 pm
cramya wrote:( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A
substitute n=1

( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) ) = (a+b)(a-b) = a^2-b^2



only A satisfies.
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Good exponent problem

by Stuart@KaplanGMAT » Thu Mar 12, 2009 11:39 pm
cramya wrote:( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A
Picking numbers is a great way to go. We could also solve this in about 15 seconds by recognizing that we have a difference of squares and peeking at the answer choices.

Any quadratic in the form of:

(x + y) (x - y)

can be rewritten as:

x^2 - y^2

Once we see that the original expression is a difference of squares, we should quickly spot that only (A) follows the form that we want.

Down with algebra! Up with critical thinking!

:D
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 160
Joined: Fri May 30, 2008 7:10 pm
Thanked: 10 times
GMAT Score:600

by dendude » Fri Mar 13, 2009 8:00 am
I have a doubt.
I recognised the pattern but I also tried to quickly solve.

a^2^(n-1) can be written as a^(2*(n-1))

Using this, I arrive at a^(4*(n-1)) - b^(4*(n-1)) and hence (a^4n)/(a^4) - (b^4n)/(b^4)
How do I use this to arrive at the answer format?
I know that this is not the most time-conscious way to solve, but I was just a little curious.
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Fri Mar 13, 2009 11:51 am
dendude wrote:I have a doubt.
I recognised the pattern but I also tried to quickly solve.

a^2^(n-1) can be written as a^(2*(n-1))

Using this, I arrive at a^(4*(n-1)) - b^(4*(n-1)) and hence (a^4n)/(a^4) - (b^4n)/(b^4)
How do I use this to arrive at the answer format?
I know that this is not the most time-conscious way to solve, but I was just a little curious.


I'm also not convinced that (a) is mathematically correct, it's just the only option in difference of squares format.

Arranging the terms slightly differently than you have (i.e. not writing as fractions), we end up with:

a^(4n- 4) - b^(4n - 4)

which is definitely not the same as:

(a) a^(2n) - b^(2n)

What's the source of the question?
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Fri Mar 13, 2009 5:33 pm
Stuart,
My bad for not including source.Its from one of the quantitative aptitude books.

My approach was this and please correct me if I am mistaken. I felt picking numbers was may be easier like u said than the algebric approach below. Posted it to point out the application of the exponent rules.


( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

Of the form (a+b) (a-b) = a^2 - b^2

a = a^2^(n-1) and b = b^2^(n-1)



( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

= (a^2^(n-1)) ^ 2 - (b^2^(n-1))^2


I then interpreted this (x^y)^z = x ^ (y*z)

where x= a ; y = 2^(n-1) ; z=2 and x=b;y = 2^(n-1) ; z=2

a ^(2^(n-1) * 2) - b ^ (2^(n-1) * 2)

= a ^(2^(n-1) * 2^1) - b ^ (2^(n-1) * 2^1)

{Using x^c * x^d = x^(c+d) where x=2,c=n-1 d=1}


= a ^ (2^n) - b ^ ( 2^n)

= a^2^n - b^2^n



I tried to represent it as unambigous as possible and in that process may have missed what I wanted to achieve.

Also my thought was if its given

as a^b^c then this is not equal to a ^(b*c)

Only (a^b)^c = a ^ (b*c)


There was also a GMAT prep that read (may not be the exact prob but just to illustrate) 2^3^4 where we had to 3^4 first then raise 2 to that power

2^3^4 is not equal to 2 ^(3*4)


Sorry for the long post....


Regards,
Cramya
Top
User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

Re: Good exponent problem

by logitech » Sat Mar 14, 2009 8:52 am
2(n-1)'s are distracting - the question stem is actually

(a^x+b^X)(a^x-b^x) which is equal to

a^2x-b^2x

I am not mistaken this is equal to: a^(4(n-1))-b^(4(n-1))
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Top
Master | Next Rank: 500 Posts
Posts: 431
Joined: Sat Jan 10, 2009 9:32 am
Thanked: 16 times
Followed by:1 members

by kanha81 » Thu Mar 26, 2009 1:35 pm
cramya wrote:
( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

Of the form (a+b) (a-b) = a^2 - b^2

a = a^2^(n-1) and b = b^2^(n-1)



( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

= (a^2^(n-1)) ^ 2 - (b^2^(n-1))^2


I then interpreted this (x^y)^z = x ^ (y*z)

where x= a ; y = 2^(n-1) ; z=2 and x=b;y = 2^(n-1) ; z=2

a ^(2^(n-1) * 2) - b ^ (2^(n-1) * 2)

= a ^(2^(n-1) * 2^1) - b ^ (2^(n-1) * 2^1)

{Using x^c * x^d = x^(c+d) where x=2,c=n-1 d=1}


= a ^ (2^n) - b ^ ( 2^n)

= a^2^n - b^2^n
I could not help, but acknowledge an excellent way to solve the problem besides choosing the number!

Thanks cramya.
Want to Beat GMAT.
Always do what you're afraid to do. Whoooop GMAT
Top
Post new topic Post Reply
• Page 1 of 1