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question 1/(2(10)^35

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by selango » Wed Nov 10, 2010 3:19 am
(1/5)^m.(1/4)^18=1/2(10)^35

m=?

(1/5)^m.(1/4)^18=1/2.(2.5)^35

(1/5)^m.(1/4)^18=1/2.2^35.5^35

(1/5)^m.(1/4)^18=1/2^36.5^35

(1/5)^m.(1/4)^18=1/4^18.5^35

(1/5)^m.(1/4)^18=1/5^35 . 1/4^18

-->m=35

Pick D

Hope this clarify!!!
--Anand--
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by [email protected] » Wed Nov 10, 2010 3:28 am
selango thank you first of all for fast reply
but I don't understand it fully

from here I dont understand:
(1/5)^m.(1/4)^18=1/2^36.5^35

(1/5)^m.(1/4)^18=1/4^18.5^35

(1/5)^m.(1/4)^18=1/5^35 . 1/4^18

how do you get the 36.5
and the 18.5
and the last one fully dont understand.
sorry for inconvenience
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by kvcpk » Wed Nov 10, 2010 3:40 am
[email protected] wrote:selango thank you first of all for fast reply
but I don't understand it fully

from here I dont understand:
(1/5)^m.(1/4)^18=1/2^36.5^35

(1/5)^m.(1/4)^18=1/4^18.5^35

(1/5)^m.(1/4)^18=1/5^35 . 1/4^18

how do you get the 36.5
and the 18.5
and the last one fully dont understand.
sorry for inconvenience
Well, I think Anand meant . to be * [multiplication]

Let me rewrite those for you:

(1/5)^m * (1/4)^18=1/(2^36 * 5^35)

(1/5)^m * (1/4)^18=(1/4)^18 * (1/5)^35

(1/5)^m * (1/4)^18=(1/5)^35 * (1/4)^18

Hence m=35.

Does this help?
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by [email protected] » Wed Nov 10, 2010 3:44 am
Well, I think Anand meant . to be * [multiplication]

Let me rewrite those for you:

(1/5)^m * (1/4)^18=1/(2^36 * 5^35)

(1/5)^m * (1/4)^18=(1/4)^18 * (1/5)^35

(1/5)^m * (1/4)^18=(1/5)^35 * (1/4)^18

Hence m=35.

Does this help?[/quote]

thank you for your help,
but the thing what I dont understand is, where do you get the 36 from and the 5 from?
and why do you do in the second one all of a sudden 1/4^18, while in the previous one it was 1/2?
sorry once again
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by Geva@EconomistGMAT » Wed Nov 10, 2010 4:21 am
https://www.beatthegmat.com/gmat-prep-ex ... 68795.html
Geva
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by MAAJ » Wed Nov 10, 2010 10:14 am
I recommend you to read these 2 topics, they are pretty good.


Exponent Manipulation: Tough Questions, Basic Approaches
https://www.beatthegmat.com/mba/2010/05/ ... approaches

The Powers That Be: Solving Tough Exponent Problems
https://www.beatthegmat.com/mba/2009/10/ ... t-problems
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by ssidda01 » Wed Nov 10, 2010 1:21 pm
[email protected] wrote:Well, I think Anand meant . to be * [multiplication]

Let me rewrite those for you:

(1/5)^m * (1/4)^18=1/(2^36 * 5^35)

(1/5)^m * (1/4)^18=(1/4)^18 * (1/5)^35

(1/5)^m * (1/4)^18=(1/5)^35 * (1/4)^18

Hence m=35.

Does this help?
Usmana..

The right hand side(RHS) of the equation is 1 / 2 (10)^35. Since 10 = 2*5 the RHS can be rewritten as 1 / 2 (2*5)^35 i.e
1 / 2 * 1 / (2)^35 * 1 / (5)^35. Now multiply 1 / 2 and 1 / (2)^35 to get 1 / (2)^36.
To get to the solution of this problem we need to equate the LHS(left hand side) of the equation to the RHS. Therefore rewrite 1 / (2)^36 as 1 / (4)^18 by splitting 1 / (2)^36 into 1 / (2)^18 * 1 / (2)^18.

To better understand the concepts above please do a chapter on Exponents.
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by ssidda01 » Wed Nov 10, 2010 1:22 pm
[email protected] wrote:Well, I think Anand meant . to be * [multiplication]

Let me rewrite those for you:

(1/5)^m * (1/4)^18=1/(2^36 * 5^35)

(1/5)^m * (1/4)^18=(1/4)^18 * (1/5)^35

(1/5)^m * (1/4)^18=(1/5)^35 * (1/4)^18

Hence m=35.

Does this help?
Usmana..

The right hand side(RHS) of the equation is 1 / 2 (10)^35. Since 10 = 2*5 the RHS can be rewritten as 1 / 2 (2*5)^35 i.e
1 / 2 * 1 / (2)^35 * 1 / (5)^35. Now multiply 1 / 2 and 1 / (2)^35 to get 1 / (2)^36.
To get to the solution of this problem we need to equate the LHS(left hand side) of the equation to the RHS. Therefore rewrite 1 / (2)^36 as 1 / (4)^18 by splitting 1 / (2)^36 into 1 / (2)^18 * 1 / (2)^18.

To better understand the concepts above please do a chapter on Exponents.
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