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Alg-exponents

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Alg-exponents [#permalink] New post 03 Dec 2008, 13:24
00:00
A
B
C
D
E

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Question Stats:

40% (01:43) correct 60% (00:01) wrong based on 5 sessions
If 5^{21} * 4^{11} = 2 * 10^n, what is the value of n?


A. 11
B. 21
C. 22
D. 23
E. 32
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Re: Alg-exponents [#permalink] New post 03 Dec 2008, 13:49
samlosco wrote:
If 5^21 * 4^11 = 2 * 10^n, what is the value of N?


a) 11
b) 21
c) 22
d) 23
e) 32


B

5^21 * 4^11
= 5^21 * (2*2)^11
= 5^21 * 2^11 * 2^11
= 5^21 * 2^22
= 5^21 * 2^21 * 2
= 2 * 10^21
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Re: Alg-exponents [#permalink] New post 03 Dec 2008, 13:59
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I agree the answer is B.

This is no different than what zoinnk did. I'm just explaning the steps taken and rules that allow us to get to the answer.

It may help someone to break down these exponents for better understanding.

If you have 4^{11}

That's the same thing as
4*4*4*4*4*4*4*4*4*4*4

Break down the 4's into 2's

(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)*(2*2)

Now count the 2's. There are 22, so 4^{11} = 2^{22}

RULE: When multiplying numbers with exponents the base numbers (2 is the base of 2^3) can be multiplied if the exponent number is the same. Here we have 5^21 and 2^22. We can pull out a 2 from 2^22 to give us 2^21 * 2 * 5^21. Now we have similar exponent: 21. 5x2 = 10, so we have 2 * 10^21. When you multiply the bases the exponent must be the same, so the exponent remains the same when finished, hence 5^{21}*2^{21} = 10^{21} We pulled out a 2 in order to make the exponent 21 rather than 22 so we have:

2 * 10^21 which makes the value of n = 21.

samlosco wrote:
If 5^21 * 4^11 = 2 * 10^n, what is the value of N?


a) 11
b) 21
c) 22
d) 23
e) 32

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Re: Alg-exponents [#permalink] New post 04 Dec 2008, 09:48
Answer is B. Here is the approach

5^21 * 4^11 = 2 * 10^n
(5^11*5^10) * (2^11*2^10*2) = 2 * 10^n (Rule is = a ^ (m+n) = a ^ m * a * n )
10^11*10^10* 2 = 2 * 10^n (Multiple the base terms with same powers. a^m * b^ m = ab ^ m)
10^21*2 = 2*10^n

n = 21
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PS: Exponents [#permalink] New post 28 Apr 2009, 09:34
If 5^{21}*4^{11} = 2*10^n what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Please explain.

I know I am suppose to work out the equation to get the integers to equal each other but just can't grasp it.

Thanks!
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Re: PS: Exponents [#permalink] New post 28 Apr 2009, 10:01
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zalan wrote:
If 5^21 X 4^11 = 2 X 10^n what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Please explain.

I know I am suppose to work out the equation to get the integers to equal each other but just can't grasp it.

Thanks!



> 5^21 * (2^2)*11 = 2 * 10^n
> 5^21 * 2^22 = 2 * (2^n*5^n)
> 5^21 * 2^22 = 2^n+1 * 5^n

i.e 5^21 = 5^n
n = 21

>2^22 = 2^n+1
>n+1 = 22.
>n=21

B
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Re: PS: Exponents [#permalink] New post 29 Apr 2009, 08:41
5^21 * 4^11 = 2 * 10^n

5^21 * 2^22 = 2 * 10^n

5^21 * 2^21= 10^n

10^21 = 10^n

n=21
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GMAT Prep 1: Exponents [#permalink] New post 13 Feb 2010, 09:52
If 5^21 x 4^11 = 2 x 10^n, what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Clear, concise explanations welcome!

[Reveal] Spoiler:
OA = B
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Re: GMAT Prep 1: Exponents [#permalink] New post 13 Feb 2010, 09:57
carriedinterest wrote:
If 5^21 x 4^11 = 2 x 10^n, what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Clear, concise explanations welcome!

[Reveal] Spoiler:
OA = B


5^21 * 4^11 = 5^21 * 2^22 = 2 * (5*2)^21 = 2 * 10^21
hence B.
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Re: GMAT Prep 1: Exponents [#permalink] New post 13 Feb 2010, 10:10
Break down the exponents on each side and the answer reveals itself:

5^{21} * 4^{11} = 2 * 10^n

Left side: 4^{11} = (2 * 2)^{11} = 2^{22}

Right side: 2 * 10^n = 2 * (5 * 2)^n = 2 * 5^n * 2^n

2^n * 2 = 2^{n+1}

Back together: 5^{21} * 2^{22} = 5^n * 2^{n + 1}

n = 21
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Re: GMAT Prep 1: Exponents [#permalink] New post 13 Feb 2010, 10:13
carriedinterest wrote:
If 5^21 x 4^11 = 2 x 10^n, what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Clear, concise explanations welcome!


5^21 x 4^11 = 2 x 10^n
5^21 x 2^22 = 2 x 10^n
(5 x 2)^21 x 2 = 2 x 10^n
10^21 x 2 = 2 x 10^n

Therefore n = 21

B
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Re: GMAT Prep 1: Exponents [#permalink] New post 13 Feb 2010, 18:57
carriedinterest wrote:
If 5^21 x 4^11 = 2 x 10^n, what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Clear, concise explanations welcome!

[Reveal] Spoiler:
OA = B


carriedinterest, I've seen this a few times - in the middle of a thread, you reprint the question from the first post, asking for 'clear, concise explanations' when there are already one or explanations that have already been provided - perhaps you did not find them clear or concise? In any case, I'm curious why you've done this; you did the same in these threads, for example:

gmatclub.com/forum/exponent-question-90368.html
gmatclub.com/forum/gmat-prep-1-symbolism-90366.html

If there is something in the solutions already given that is unclear to you, then it would be more helpful to point that out; surely other test takers share your concerns, and you can all benefit from having that detail of the solution explained. Or, if you are asking if there are alternative solution methods, it would be good to mention that. Otherwise, as has happened in each of these threads, people simply end up typing up the same solutions that were posted above, and I don't know that that benefits anyone much.
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Re: GMAT Prep 1: Exponents [#permalink] New post 14 Feb 2010, 02:22
Expert's post
IanStewart wrote:
carriedinterest wrote:
If 5^21 x 4^11 = 2 x 10^n, what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Clear, concise explanations welcome!

[Reveal] Spoiler:
OA = B


carriedinterest, I've seen this a few times - in the middle of a thread, you reprint the question from the first post, asking for 'clear, concise explanations' when there are already one or explanations that have already been provided - perhaps you did not find them clear or concise? In any case, I'm curious why you've done this; you did the same in these threads, for example:

gmatclub.com/forum/exponent-question-90368.html
gmatclub.com/forum/gmat-prep-1-symbolism-90366.html

If there is something in the solutions already given that is unclear to you, then it would be more helpful to point that out; surely other test takers share your concerns, and you can all benefit from having that detail of the solution explained. Or, if you are asking if there are alternative solution methods, it would be good to mention that. Otherwise, as has happened in each of these threads, people simply end up typing up the same solutions that were posted above, and I don't know that that benefits anyone much.


Ian, It was I, who merged the topics as carriedinterest posted several question that were already discussed. Guess I should mentioned this in the thread after doing so. Sorry for confusion.
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Re: Alg-exponents [#permalink] New post 14 Feb 2010, 03:26
Ah, I see - my apologies! It did seem very odd, so I was curious to know why it was happening, but that makes a lot of sense.
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Re: Alg-exponents [#permalink] New post 05 Feb 2011, 21:40
These types are usually solved by reducing the term to prime factors and few common rules of exponent.

5^21 * 4^11 = 2 * 10^n

Consider the bases first; 5,4,2,10
5 and 2 are already primes; 4 and 10 composites.
Reduce these to prime factors:
4=2*2
10=5*2
Place in the equation:

5^{21} * 4^{11} = 2 * 10^n
5^{21} * (2*2)^{11} = 2 * (2*5)^n
5^{21} * (2)^{11} *(2)^{11}= 2 * (2)^n*(5)^n
##(x*y)^m = x^m * y^m#### Rule of exponent##
5^{21} * (2)^{22} = (5)^n * 2^{(n+1)}
##x^m*x^n = x^{(m+n)}## Rule of exponent##

Compare LHS and RHS
n+1=22
n=21

Ans: "B"
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Re: Alg-exponents   [#permalink] 05 Feb 2011, 21:40
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