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OG Problem Solving Question #11

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by acrowson » Wed Feb 03, 2010 12:52 pm
Hey guys,
I am trying to figure out the solution (pg. 191) for question #11 in the problem solving section.

Specifically, how the heck do I get from cube root (10^-6) to 10^-2??

Any help would be appreciated!

Thanks,
Andrew
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by Brent@GMATPrepNow » Fri Feb 05, 2010 8:10 am
acrowson wrote:Hey guys,
I am trying to figure out the solution (pg. 191) for question #11 in the problem solving section.

Specifically, how the heck do I get from cube root (10^-6) to 10^-2??

Any help would be appreciated!

Thanks,
Andrew
The important concept here is that cuberootx = x^(1/3)
In fact, the nth root of x is equal to x^(1/n)

So, cuberoot(10^-6) = (10^-6)^(1/3)
Applying the Power of a Power Rule, we multiply the exponents to get (10^-6)^(1/3) = 10^(-2)
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by shashank.ism » Fri Feb 05, 2010 9:05 am
acrowson wrote:
Hey guys,
I am trying to figure out the solution (pg. 191) for question #11 in the problem solving section.

Specifically, how the heck do I get from cube root (10^-6) to 10^-2??

Any help would be appreciated!

Thanks,
Andrew
its very simple as illustrated by, Brent Hanneson:
you can also understand it in a different way as below:
we can write 10^-6 = 1/10^6
cube root of 10^-6 = (10^-6)^(1/3) = (1/10^6)^(1/3) = 1/((10^6)^(1/3) ) -----(i)


now as u had confusion in calculating cube root of 10^-6 which is either way possible..
so I had simplified to a term which contains cube root of 10^6. Now I think u understand that cube root of 10^6 = 10^2

so from (i) cube root of 10^-6 = 1/((10^6)^(1/3) ) = 1/10^2 = 10^-2

I hope you also understand how 1/10^2 = 10^-2.


Well for clearing your concept completely you should refer a book of algebra , chapter power and indices. you can easily understand the concept.
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by acrowson » Fri Feb 05, 2010 9:42 am
Thanks guys for the help...
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