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Good exponent problem

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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.
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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
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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.
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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?
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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
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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))
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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.
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