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Exponent Problem - Kinda Tricky!

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by student22 » Wed Apr 14, 2010 4:32 pm
I used the equation editor to make the solution as clear as possible. Basically what I did was multiply each of the 3^7 and 3^8 exponents by "1" in order to convert the denominators into the same exponent. Hopefully, it's clear, if not let me know.
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The --> arrows just mean, next step...
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by pkw209 » Wed Apr 14, 2010 4:37 pm
That's what I got as well. Pulled this from another source so I thought the answer was E.

Fantastic job and explanation. That type of effort is what makes this site awesome.
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by Fiver » Wed Apr 14, 2010 7:12 pm
pkw209 wrote:Hey guys,

Couldn't figure this one out. An explanation would be great! Thanks.

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Another way is to apply the formula for the total sum in a GP.

x - (2/3^7 + 2/3^8 + 2/3^9) = 1/3^9
the bold part is a GP with FT as 2/3^9 and common ration as 3
Sum of a GP = FT * (r^n - 1)/ r-1 )
sum = 2/3^9 ( 3^3 - 1) / (3-1)
= 2/3^9 *(26/2) = 26/3^9

Therefore x - 26/3^9 = 1/3^9
x = 27/3^9
x = 1/3^6
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by kstv » Wed Apr 14, 2010 8:24 pm
Multiply by 3^8 it is kinda of a middle value , which makes the calc manageable
3^8X - 2*3 - 2 - 2/3 = 1/3
3^8 X = 6+2+1 = 3²
X = 1/3^6
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by thephoenix » Thu Apr 15, 2010 3:20 am
let 3^7=y
the eqn becomes x-2/y-2/3y-2/9y=1/9y
simplifying
(9xy-18-6-2)/9y=1/9y
or
9xy-18-6-2=1 or
9xy=27
or
x=27/9*3^7=3/3^7=1/3^6
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