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(1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 08:22
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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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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Updated on: 23 Aug 2019, 09:12
IMOB
(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.



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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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Updated on: 23 Aug 2019, 09:34
\((1/3)^{3  2*2  1*3 +0} = (1/3)^{10}\) Answer is (B)
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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.



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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 08:32
(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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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 08:40
Quote: \((1/3)^{−3}(1/9)^{−2}(1/27)^{−1}(1/81)^0=\)
\(A. (1/3)^{−36}\)
\(B. (1/3)^{−10}\) \(C. (1/3)^{−11}\)
\(D. (1/9)^{−10}\)
\(E. (1/9)^{11}\) \((1/3)^{−3}(1/9)^{−2}(1/27)^{−1}(1/81)^0 = (1/3)^{−3}(1/3)^{−4}(1/3)^{−3}(1/3)^0 = (1/3)^{343+0} =(1/3)^{10}\) IMO B
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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 08:43
This is how I did. Hope it helps. Posted from my mobile device
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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 08:55
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



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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 11:02
\((\frac{1}{3})^−3\)\((\frac{1}{9})^−2\)\((\frac{1}{27})^−1\)\((\frac{1}{81})^0\) = 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 \((\frac{1}{3})^−3\)\((\frac{1}{9})^−2\)\((\frac{1}{27})^−1\)\((\frac{1}{81})^0\) \((\frac{1}{3})^−3\)\((\frac{1}{3})^−4\)\((\frac{1}{3})^−3\)\((\frac{1}{3})^0\) \(((\frac{1}{3})^(−34−3))\)*1 \((\frac{1}{3})^10\) Answer (B).
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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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23 Aug 2019, 11:33
\((1/3)^{3} (1/9)^{2} (1/27)^{1} (1/81)^0=\) \(=(1/3)^{3} (1/3)^{4} (1/3)^{3}\) since \((1/81)^0=1\) \(=(1/3)^{10}\) Answer (B)
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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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24 Aug 2019, 02:05
(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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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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24 Aug 2019, 02:44
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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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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24 Aug 2019, 03:40
\((\frac{1}{3})^{−3}* (\frac{1}{9})^{−2}* (\frac{1}{27})^{−1}* (\frac{1}{81})^{0}\) =
= \((\frac{1}{3})^{−3}* (\frac{1}{3})^{2*(2)}* (\frac{1}{3})^{(1)*3}* 1\) =
= \((\frac{1}{3})^{3}* (\frac{1}{3})^{4}* (\frac{1}{3})^{3}\) =
= \((\frac{1}{3})^{343}= (\frac{1}{3})^{10}\)
The answer is B.



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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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24 Aug 2019, 11:51
\((\frac{1}{3})^{3}(\frac{1}{9})^{2}(\frac{1}{27})^{1}(\frac{1}{81})^0=\)
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 \((1/3)^{10}\)
Answer B



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Re: (1/3)^(3)(1/9)^(2)(1/27)^(1)(1/81)^0
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25 Aug 2019, 02:49
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




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