Author
Message
CEO

Joined: 21 Jan 2007

Posts: 2756

Location: New York City

Followers: 9

Kudos [? ]:
773
[0 ] , given: 4

Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
12 Nov 2007, 09:11

2

This post was BOOKMARKED

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The square of \(5^{\sqrt{2}}\) = ?

(A) \(5^2\)

(B) \(25^{\sqrt{2}}\)

(C) \(25\)

(D) \(25^{2\sqrt{2}}\)

(E) \(5^{\sqrt{2}^2}\)

Source: GMAT Club Tests - hardest GMAT questions

SOLUTION: challenge-radical-exponents-m06q20-55442-20.html#p1232318

CIO

Joined: 02 Oct 2007

Posts: 1218

Followers: 95

Kudos [? ]:
883
[5 ]
, given: 334

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
27 May 2010, 23:49
5

This post received KUDOS

Here's a quote from the thread mentioned above:

Quote:

Tania, consider these examples: \((5^1)^2 = 5^2 = 25\) --> the answer is \(5^2\), not \(25^2\), which would equal \(5^4\) (incorrect). \((5^2)^2 = 5^{2*2} = 5^4 = 625 = 25^2\) --> You see that we had to multiply the exponents (2*2) but didn't change the base at that stage yet. If we follow your logic we end up with \(25^{2*2}\), which is not right since we've squared the expression \(5^2\) twice, not once (we squared the base and multiplied the exponent by 2). Let's see our problem again: \((5^{\sqrt{2}})^2 = 5^{2\sqrt{2}} = (5^2)^{\sqrt{2}} = 25^{\sqrt{2}}\) --> make sure you square the expression once So, the right answer could be either \(25^{\sqrt{2}}\) or \(5^{2\sqrt{2}}\). I hope it helped make it a bit clearer.

_________________

Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.GMAT Club Premium Membership - big benefits and savings

Director

Joined: 21 Dec 2009

Posts: 591

Concentration: Entrepreneurship, Finance

Followers: 18

Kudos [? ]:
629
[3 ]
, given: 20

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
27 May 2010, 07:59
3

This post received KUDOS

5^\sqrt{2} x 5^\sqrt{2}

=> 5^\sqrt{2}+\sqrt{2}

=> 5^2\sqrt{2}

= 25^\sqrt{2}

B is correct.

_________________

KUDOS me if you feel my contribution has helped you.

CIO

Joined: 02 Oct 2007

Posts: 1218

Followers: 95

Kudos [? ]:
883
[1 ]
, given: 334

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
10 Jun 2011, 04:31
1

This post received KUDOS

You should see this thread:

writing-mathematical-symbols-in-posts-72468.html We can display quite sophisticated mathematical expressions in our forum, like this:

\(\sqrt{(x^2 - 2x +1)^{\frac{2}{\sqrt{3}}}}\)

arunangsude2011 wrote:

hi, i was struggling to write ' to the power ' in symbols. can somebody guide me please. thanks

_________________

Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.GMAT Club Premium Membership - big benefits and savings

Math Forum Moderator

Joined: 20 Dec 2010

Posts: 2021

Followers: 158

Kudos [? ]:
1592
[1 ]
, given: 376

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
10 Jun 2011, 04:32
1

This post received KUDOS

arunangsude2011 wrote:

hi, i was struggling to write ' to the power ' in symbols. can somebody guide me please. thanks

a to the power b:

a^b, ^ is called a cap and typed using "shift 6" in most keyboards.

m tag:

\(a^b\)

-1 to the power m+n

(-1)^(m+n)

m tag:

\((-1)^{(m+n)}\)

_________________

~flukeGMAT Club Premium Membership - big benefits and savings

Senior Manager

Joined: 01 Sep 2006

Posts: 301

Location: Phoenix, AZ, USA

Followers: 1

Kudos [? ]:
24
[0 ] , given: 0

square of 5^(root2)=
(5^(root2)^2==
5^(root2)^2

Director

Joined: 09 Aug 2006

Posts: 763

Followers: 1

Kudos [? ]:
168
[0 ] , given: 0

Re: Challenge - Radical Exponents [#permalink ]

Show Tags
12 Nov 2007, 11:13

bmwhype2 wrote:

The square of 5^sqrt(2) is 5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2 Can someone please explain the answer?

I'm getting B.

[5^sqrt(2)]^2 = (5^2)^sqrt(2) = 25^sqrt(2)

Director

Joined: 25 Oct 2006

Posts: 648

Followers: 12

Kudos [? ]:
458
[0 ] , given: 6

I couldn't find the answer in the given stem.
9.75 is caculated value and none of the answer has that output
5^2 = 25
25^sqrt(2) = 94.77
25
25^2sqrt(2) = 46.08
5^sqrt(2)^2 = 25

Manager

Joined: 08 Nov 2007

Posts: 99

Followers: 1

Kudos [? ]:
2
[0 ] , given: 0

I get 5^2
I tried substituting 4 for 2 - (5^sqrt4) x (5^sqrt4)
5^sqrt4 = 5^2 = 25 - so 25 x 25 = 625 = 5^4
On that basis I'd say (5^sqrt2) x (5^sqrt2) = 5^2
Anyone with views?

Current Student

Joined: 18 Jun 2007

Posts: 408

Location: Atlanta, GA

Schools: Emory class of 2010

Followers: 11

Kudos [? ]:
40
[0 ] , given: 0

Re: Challenge - Radical Exponents [#permalink ]

Show Tags
13 Nov 2007, 11:43

bmwhype2 wrote:

The square of 5^sqrt(2) is 5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2 Can someone please explain the answer?

I get B, 25^sqrt(2)

When an exponent is raised to another exponent, they are multiplied. Thus,

The square of 5^sqrt(2) = 5^(2*sqrt(2)), then calculating backwards,

5^(2*sqrt(2)) also equals [5^(2)]^sqrt(2) = [25]^sqrt(2)

Manager

Joined: 19 Aug 2007

Posts: 169

Followers: 1

Kudos [? ]:
44
[0 ] , given: 0

priyankur just to let u knwo i figured out what we did wrong. we didnt square the original statement
5^(sqrt2) = 9.738 but then we have to square that to get 94.838

CIO

Joined: 02 Oct 2007

Posts: 1218

Followers: 95

Kudos [? ]:
883
[0 ] , given: 334

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
22 Dec 2009, 02:39

Intern

Joined: 27 May 2010

Posts: 1

Followers: 0

Kudos [? ]:
0
[0 ] , given: 0

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
27 May 2010, 13:54

The square of 5^sqrt(2) is 5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2 Can someone please explain the answer? B is the answer [5^sqrt(2)]^2 = 25^2sqrt(2)

Intern

Joined: 27 May 2010

Posts: 6

Followers: 0

Kudos [? ]:
23
[0 ] , given: 0

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
28 May 2010, 03:06

Answer is 25^root2 2^2^3 is same as 4^3

CIO

Joined: 02 Oct 2007

Posts: 1218

Followers: 95

Kudos [? ]:
883
[0 ] , given: 334

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
28 May 2010, 03:52

Yes, the answer is B. However, your statement in red is not correct.

\(2^{2^3} = 2^8 = 256\)

\(4^3 = 64\)

If you meant to type \(2^{2*3}\) is same as \(4^3\), then you're correct.

ameyaberi wrote:

Answer is 25^root22^2^3 is same as 4^3

_________________

Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.GMAT Club Premium Membership - big benefits and savings

Intern

Joined: 03 Aug 2010

Posts: 2

Followers: 0

Kudos [? ]:
1
[0 ] , given: 0

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
06 Aug 2010, 06:36

(5^sqrt2)^2 = 5^2^1/2*2^1....since same base we have to add the exponents = 5^2^3/2 = 5^sqrt2^3 = 5^2*sqrt2 = 25^sqrt2...Thus B

SVP

Joined: 16 Nov 2010

Posts: 1673

Location: United States (IN)

Concentration: Strategy, Technology

Followers: 34

Kudos [? ]:
481
[0 ] , given: 36

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
31 May 2011, 05:06

5^(root(2)) * 5^(root(2)) = 5^2(root(2)) = 25^(root(2))

Answer - B

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)GMAT Club Premium Membership - big benefits and savings

Intern

Joined: 29 May 2011

Posts: 1

Followers: 0

Kudos [? ]:
0
[0 ] , given: 0

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
31 May 2011, 08:50

answer is B (a^m)^n = a^mn = (a^n)^m hence (5^sqrt2)^2 = 5^sqrt2*2 = (5^2)^sqrt2 = 25^sqrt2 hope that helps.

Manager

Joined: 19 Apr 2011

Posts: 111

Followers: 2

Kudos [? ]:
3
[0 ] , given: 2

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
01 Jun 2011, 06:47

I get the following results.

Attachments

explantionfor5root.JPG [ 17.97 KiB | Viewed 7321 times ]

CIO

Joined: 02 Oct 2007

Posts: 1218

Followers: 95

Kudos [? ]:
883
[0 ] , given: 334

Re: Challenge - Radical Exponents (m06q20) [#permalink ]

Show Tags
01 Jun 2011, 07:19

As it was correctly stated by abcg27 (you can see the quote below), you've taken the same path. If you go one step further, you'll arrive at B:

\(5^{2\sqrt{2}} = (5^2)^{\sqrt{2}} = 25^{\sqrt{2}}\)

abcg27 wrote:

answer is B (a^m)^n = a^mn = (a^n)^m hence (5^sqrt2)^2 = 5^sqrt2*2 = (5^2)^sqrt2 = 25^sqrt2 hope that helps.

toughmat wrote:

I get the following results.

_________________

Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.GMAT Club Premium Membership - big benefits and savings

Re: Challenge - Radical Exponents (m06q20)
[#permalink ]
01 Jun 2011, 07:19

Go to page
1 2
Next
[ 28 posts ]