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Re: Very confused [#permalink]
@phoenixio

thanks a lot, stupid of me...
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Re: Very confused [#permalink]
phoenixio wrote:
@Sonalika,
in your 2nd case a-b=-4 => b= a+4
therefore b>a

I would have solved this way

\(3^{a-b}\)= \(\frac{1}{81}\)
=> \(\frac{1}{3^{b-a}}\) = \(\frac{1}{81}\)
therefore b>a

Does this mean the answer is B? The question still carries the answer as D. I mean, it is intuitive that is a-b = -4, then a has to be less than b for it to be negative, right?
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
@emike56

yup,you're right.
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
1
ste133 wrote:
@emike56

yup,you're right.


@carcass please change the answer to show B and not D. Otherwise, please provide the working solution that suggests its D.
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
Expert Reply
Fixed. Thank you.
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
Okay, this is what I did. 1/81 = 1/(3^4)= 3^-4.
a-b=-4 which means B should be greater. Is this approach correct?
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
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Correct approach :

3^(a-b) = 1/81
=3^(a-b) = 3^(-4)
comparing bases,
a-b = -4
hence, a = b - 4
which proves, a is 4 units smaller than b , no matter what.
Hence, B.

Cheers !
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
1
Answer: B

3^(a-b) = 1/81
3^(a-b) = 3^-4
a - b = -4
b - a = 4 and because b-a is positive, b is bigger than a.
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Re: Given that (3)^(a-b) = 1/81 [#permalink]
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