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walker
(3^(k-1))^3=(3^3k)*(3^(-3))=root((3^6k))*(3^(-3))=root(8100)*1/27=90/27=10/3

i follow everything until the second equal sign. Why are you equating the two expressions ?


If I understand you right I'm not equating.
do you ask me about (3^3k)=root(3^6k) ?
I made this to put 3^6k from the equation into the expression without knowing k.
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(3^(k-1))^3=(3^3k)*(3^(-3))=root((3^6k))*(3^(-3))=root(8100)*1/27=90/27=10/3


so smoth like your biking.
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(3^(k-1))^3=(3^3k)*(3^(-3))=root((3^6k))*(3^(-3))=root(8100)*1/27=90/27=10

walker, i like the fact that you managed to solve this without even looking for K, and that can help save time. However, I don't understand the area that i've bolded. Why did you root that expression? maybe I don't understand the rules and so would need your input, but I don't see how root(3^6k) can equal to 3^3k. would you please explain? thanks
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GMAT TIGER
walker
(3^(k-1))^3=(3^3k)*(3^(-3))=root((3^6k))*(3^(-3))=root(8100)*1/27=90/27=10/3

so smoth like your biking.

:-D Thanks Tiger! I really like race! but sometimes it was not so smooth as equations :-D
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I don't see how root(3^6k) can equal to 3^3k. would you please explain? thanks


m^n=m^(n*2*1/2)=(m^(2*n))^1/2
or general rule: x^(n*m)=(x^n)^m=(x^m)^n

for example,

16=2^4=2^(4*2*1/2)=2^(8*1/2)=(2^8)^1/2=root(2^8)=root(256)=16

is it ok?
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walker
(3^(k-1))^3=(3^3k)*(3^(-3))=root((3^6k))*(3^(-3))=root(8100)*1/27=90/27=10/3



That is one sweet solution. I was wasting time trying to find the value of k.
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wow, i'm amazed! thanks a lot for the explanation walker. this approach is really something! :)



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