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question 1/(2(10)^35

Expert replies
Source: — Problem Solving |

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?
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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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
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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.
Join the discussion