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Can't seem to figure this question out

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by ssmiles08 » Fri Jun 12, 2009 4:13 am
27 = 3^3

(3^3)^n = (3^2)^4

(3)^3n = 3^8

n = 8/3
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I am not sure if it is correct

by uzbek » Sat Jun 13, 2009 11:38 pm
27^n=9^4

27 is 3 times bigger then 9
hence, n must be 3 bigger then 4
so
n = 12

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Re: I am not sure if it is correct

by 2007fall » Sun Jun 14, 2009 6:16 am
Think I did it the way ssmiles08 did ...
uzbek wrote:27^n=9^4

27 is 3 times bigger then 9
hence, n must be 3 bigger then 4
so
n = 12

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Re: I am not sure if it is correct

by ssmiles08 » Sun Jun 14, 2009 7:28 am
uzbek wrote:27^n=9^4

27 is 3 times bigger then 9
hence, n must be 3 bigger then 4
so
n = 12

There is no way 27^12 = 9^4....It even logically absurd. YOu need to convert them all into 3's since they are all multiples of 3 and then set them equal to each other.
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by uzbek » Sun Jun 14, 2009 9:04 am
Ups...
Sorry, I didn't realize that this symbol meant "^" exponent
I though it was one of those GMAT questions where they give you some weird symbols like * or @ or something
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by mccollud » Tue Jun 23, 2009 6:47 pm
Here's how I did it.

9^4 = 9*9*9*9=3*3*3*3*3*3*3*3 aka 3^8

since 27=3*3*3, i found "n" by finding how many 27's there were in 3^8

so, (3*3*3)(3*3*3)(3*3) means there 2 and 2/3 27's in 3^8

27^2.666666666=6561=9^4

-Hope this helps
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Re: Can't seem to figure this question out

by pops » Tue Jun 23, 2009 9:24 pm
mets3145 wrote:If 27^n = 9^4

what's n?

how do i isolate n?
Normally in such problems simplify the base to the lowest possible number, then its easier to solve.

27^n = 9^4
3^3n = 3^8
3n=8
n=8/3
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