Difficult Exponents Problem

Expert replies

mattocks: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sun Mar 07, 2010 1:17 pm

by mattocks » Sun Jul 11, 2010 8:50 am

If x=10^100 and x^x=10^k, k=?.

A) 10^5
B) 10^102
C) 10^99
D) 10^200
E) 10^98

I'm really not sure how to attack this problem. Any suggestions would be greatly appreciated.

Like this post

Source: Beat The GMAT — Problem Solving | Sun Jul 11, 2010 8:50 am

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Sun Jul 11, 2010 9:13 am

x=10^100 --Eqn1

x^x=10^k --Eqn 2

We can solve it in 2 ways.

Substitute x value in eqn2

(10^100)^(10^100)=10^k

10^(100*10^100)=10^k[as (a^b)^c=a^bc]

10^(10^2*10^100)=10^k

10^(10^102)=10^k

k=10^102

-------------------------------------------------

Another approach

x=10^100 --Eqn1

x^x=10^k --Eqn 2

Raise to power x on both sides on eqn 1

x^x=10^100x

-->10^100x=10^k

k=100x=100*10^100=10^2*10^100=10^102

Hope this clarify

--Anand--

Like this post

mattocks: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sun Mar 07, 2010 1:17 pm

by mattocks » Sun Jul 11, 2010 9:24 am

Great explanation. Thank you ! I have one more...

If x and j are integers and (12^x)*(4^2x+1)=(2^j)*(3^2), what is the value of j?

A) 5
B) 7
C) 10
D) 12
E) 14

So 4^2x+1 = (2^2)^2x+1 = 2^4x+2. Can set 4x+2 = j. I'm not sure what do do with 12^x. Any suggestions?

Like this post

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Sun Jul 11, 2010 9:33 am

(12^x)*(4^2x+1)=(2^j)*(3^2)

12^x can be written as 3^x*4^x

3^x*4^x*(4^2x+1)=(2^j)*(3^2)

3^x*4^(3x+1)=(2^j)*(3^2)

3^x*2^(6x+2)=(3^2)*(2^j)

comparing both sides we get,

x=2 and 6x+2=j

12+2=j; j=14

Option E

--Anand--

Like this post

mattocks: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sun Mar 07, 2010 1:17 pm

by mattocks » Sun Jul 11, 2010 10:45 am

If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?

(1) The largest number of the three distinct numbers is 12.
(2) The average of the three numbers is 20/3

(1) is sufficient. If the largest number is 12, then the product of the smaller two numbers must also be 12.

(2) I tried picking some numbers for this but I couldn't find a more robust approach.

Any suggestions?

Like this post

gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sun Jul 11, 2010 11:07 am

mattocks wrote:If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?

(1) The largest number of the three distinct numbers is 12.
(2) The average of the three numbers is 20/3

(1) is sufficient. If the largest number is 12, then the product of the smaller two numbers must also be 12.

(2) I tried picking some numbers for this but I couldn't find a more robust approach.

Any suggestions?

st 1 is sufficient ..Agreed!

coming to st 2 :

a+b+c = 20

(2,6,12) & 3,4,12) both these pairs will be suffice. But for the sum of a,b,c = 20, only (2,6,12) fits in.

So st 2 is also sufficient.

since the stem asks only find the product of other 2 smaller numbers, pick D.

If they would have asked what are the other 2 smaller numbers I would have picked C.

R&D on GMAT algorithm:

https://www.beatthegmat.com/r-d-on-gmat- ... tml#305739

Like this post

barcebal: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Wed Jul 07, 2010 2:21 pmFollowed by: 2 members; Thanked: 12 times; GMAT Score:760

by barcebal » Sun Jul 11, 2010 9:26 pm

Mattocks,

To help you come up with a way to choose a, b, and c values for statement two consider this

if sqrt(abc)=c, A*B=C

Now instead of picking the A and B values, pick the C value. I'll choose 10.

A*B=10

My only options for A and B are 1, 10 and 2, 5. Can't be 1, 10 because they have to be distinct. I go with 2,5, but then A+B+C=17 so 10 doesn't work. They are less than 20 so we need C to be greater (if you are unsure whether C MUST be greater, I recommend that you start even lower, like at c=6 so that you can feel the patterns. I started at C=3 just to get a sense of what the pattern was and quickly moved up to 8.)

Try C=11
A*B=11 A and B can only be 1,11 which doesn't work

C=12
A*B=12 A and be can be 1,12; 2,6; 3;4. First set doesn't sum to, second set does, third set doesn't.

We have an answer.

I tried c=13 and c=14 to be safe. C=13 can't work because a,b,c must be unique and with c=14 a,b must be 7 and 2 which is now GREATER than 20 so C>14 is not going to work.

A*

Like this post

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Mon Jul 12, 2010 4:38 am

mattocks wrote:If x=10^100 and x^x=10^k, k=?.

A) 10^5
B) 10^102
C) 10^99
D) 10^200
E) 10^98

I'm really not sure how to attack this problem. Any suggestions would be greatly appreciated.

Another way to tackle this is using logarithm (even though it is out of scope for the GMAT, it does make life a lot easier when exponents are used). For instance

x = 10^100
log x = 100log10 = 100 (log10=1)

x^x = 10^k
xlogx = klog10 = k
=> 100x = k

Now, x = 10^100
=> 100 x (10^100) = k
=> 10^2 x 10^100 = k
=> 10^102 = k

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review