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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   15% (low)

Question Stats: 81% (01:09) correct 19% (02:08) wrong based on 78 sessions

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Competition Mode Question

$$(\frac{1}{3})^{-3}(\frac{1}{9})^{-2}(\frac{1}{27})^{-1}(\frac{1}{81})^0=$$

A. $$(\frac{1}{3})^{-36}$$

B. $$(\frac{1}{3})^{-10}$$

C. $$(\frac{1}{3})^{-11}$$

D. $$(\frac{1}{9})^{-10}$$

E. $$(\frac{1}{9})^{-11}$$

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Manager  G
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IMO-B

(1/3)^ −3 * (1/9)^−2 *(1/27)^−1 * (1/81)^0=

(1/3)^ −3 *(1/3)^ −2x2 * (1/3)^ −3x1 *(1/3)^ 0 =

1/3 ^ (-10)

Ans B

Originally posted by MayankSingh on 23 Aug 2019, 08:25.
Last edited by MayankSingh on 23 Aug 2019, 09:12, edited 1 time in total.
Senior Manager  P
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$$(1/3)^{-3 - 2*2 - 1*3 +0} = (1/3)^{-10}$$

Posted from my mobile device

Originally posted by chondro48 on 23 Aug 2019, 08:29.
Last edited by chondro48 on 23 Aug 2019, 09:34, edited 4 times in total.
GMAT Club Legend  D
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1
(13)−3(19)−2(127)−1(181)0=(13)−3(19)−2(127)−1(181)0=

A. (1/3)^−36

B. (1/3)^−10

C. (1/3)^−11

D. (1/9)^−10

E. (1/9)^−11

given expression can be written as ;
27*81*27*1 ;
IMO B ; (1/3)^−10
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Joined: 03 Jun 2019
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2
Quote:
$$(1/3)^{−3}(1/9)^{−2}(1/27)^{−1}(1/81)^0=$$

$$A. (1/3)^{−36}$$

$$B. (1/3)^{−10}$$

$$C. (1/3)^{−11}$$

$$D. (1/9)^{−10}$$

$$E. (1/9)^{-11}$$

$$(1/3)^{−3}(1/9)^{−2}(1/27)^{−1}(1/81)^0 = (1/3)^{−3}(1/3)^{−4}(1/3)^{−3}(1/3)^0 = (1/3)^{-3-4-3+0} =(1/3)^{-10}$$

IMO B
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This is how I did. Hope it helps.

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Attachments IMG_20190823_211136.jpg [ 2.24 MiB | Viewed 888 times ]

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Joined: 15 Aug 2017
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1
IMO, Flip the fraction and change the negative to positive

so:
(1/3)^(-3) = 3^3
(1/9)^(-2) = 9^2 = (3^2)^2 = 3^4
(1/27)^(-1) = 27^1 = (3^4)^1 = 3^3
(1/81)^(0) = 81^0 = (3^4)^0 = 3^0 = 1

since bases are all the same (except for the 1) add up the exponents for all base 3:
3+4+3 = 10
3^10

Change back into a fraction and turn it negative

(1/3)^-10

Ans B
Senior Manager  G
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$$(\frac{1}{3})^−3$$$$(\frac{1}{9})^−2$$$$(\frac{1}{27})^−1$$$$(\frac{1}{81})^0$$ =

A. $$(\frac{1}{3})^−36$$

B. $$(\frac{1}{3})^−10$$

C. $$(\frac{1}{3})^−11$$

D. $$(\frac{1}{9})^−10$$

E. $$(\frac{1}{9})^−11$$

Solving $$(\frac{1}{3})^−3$$$$(\frac{1}{9})^−2$$$$(\frac{1}{27})^−1$$$$(\frac{1}{81})^0$$
 $$(\frac{1}{3})^−3$$$$(\frac{1}{3})^−4$$$$(\frac{1}{3})^−3$$$$(\frac{1}{3})^0$$
 $$((\frac{1}{3})^(−3-4−3))$$*1
 $$(\frac{1}{3})^-10$$

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1
$$(1/3)^{-3} (1/9)^{-2} (1/27)^{-1} (1/81)^0=$$

$$=(1/3)^{-3} (1/3)^{-4} (1/3)^{-3}$$ since $$(1/81)^0=1$$

$$=(1/3)^{-10}$$

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Senior Manager  G
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(1/3)^-3 * (1/9)^-2 * (1/27)^-1 * (1/81)^0
= (1/3)^-3 * (1/3)^-4 * (1/3)^-3 * 1
= (1/3)^-10
Hence B.

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1
Understanding
1) When we multiply same number we add the powers -> $$(3^3)(3^2) = (3^5)$$
2) With numbers to the negative power, put the number to the given power and then flip it over -> e.g. $$3^{-1}$$ = $$\frac{1}{3}$$. Note you can think of 3 here as (3/1) to more clearly see what is going on. Similarly, $$(\frac{1}{3})^{-2}$$ = 9
3) Any number to power of zero is 1 -> $$(3^0)$$ = 1

Solution:
Now, let's breakdown the prompt bit by bit.

$$(\frac{1}{3})^{-3}$$ = $$3^3$$
$$(\frac{1}{9})^{-2}$$ = $$9^2 => 3^4$$
$$(\frac{1}{27})^{-1}$$ = 27 => $$3^3$$
$$(\frac{1}{81})^{0}$$ = 1

Thus, we get $$(3^3)(3^4)(3^3)$$ which is $$3^{10}$$. Looking at the answers, all of them use fractions. Therefore we express $$3^{10}$$ as fraction to get $$(1/3)^{-10}$$.

B is the answer.
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$$(\frac{1}{3})^{−3}* (\frac{1}{9})^{−2}* (\frac{1}{27})^{−1}* (\frac{1}{81})^{0}$$ =

= $$(\frac{1}{3})^{−3}* (\frac{1}{3})^{2*(-2)}* (\frac{1}{3})^{(-1)*3}* 1$$ =

= $$(\frac{1}{3})^{-3}* (\frac{1}{3})^{-4}* (\frac{1}{3})^{-3}$$ =

= $$(\frac{1}{3})^{-3-4-3}= (\frac{1}{3})^{-10}$$

The answer is B.
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$$(\frac{1}{3})^{-3}(\frac{1}{9})^{-2}(\frac{1}{27})^{-1}(\frac{1}{81})^0=$$

Four Exponents rules to note
1. (1/x)^-5=(x)^5 => negative exponents in the numerator get moved to the denominator and become positive exponents
2. (x)^1 => any number(except 0) to the power zero is 1,
3. $$(b^2)^3 = b^6$$ to raise a power to a power you need to multiply the exponents
4. $$x^1*x^1=x^2$$

=>$$(3^3)*(3^4)*(3^3)*1$$ =>$$3^{10}$$ or $$(1/3)^{10}$$

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convert everything in the powers of 3 since answers can easily be deduced from that
= $$\frac{1}{(3^{-3} * 3^{-4} * 3^{-3})}$$
=$$\frac{1}{3 ^ {-(3+4+3+)}}$$
= $$\frac{1}{3^{-10}}$$
= B Re: (1/3)^(-3)(1/9)^-(2)(1/27)^(-1)(1/81)^0   [#permalink] 25 Aug 2019, 02:49
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