(1/3)^(-3)(1/9)^-(2)(1/27)^(-1)(1/81)^0

Question

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}

MayankSingh · Answer

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

freedom128 · Answer

(1/3)^{-3 - 2*2 - 1*3 +0} = (1/3)^{-10}  
Answer is (B)

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Archit3110 · Answer

(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

Kinshook · Answer

(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

EncounterGMAT · Answer

This is how I did. Hope it helps.

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Kalubalu · Answer

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

unraveled · Answer

(\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

Answer (B).

Eternal · Answer

(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)

eakabuah · Answer

(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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Xin Cho · Answer

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.

lacktutor · Answer

(\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.

AvidDreamer09 · Answer

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

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

globaldesi · Answer

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

(1/3)^(-3)(1/9)^-(2)(1/27)^(-1)(1/81)^0

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(1/3)^(-3)(1/9)^-(2)(1/27)^(-1)(1/81)^0

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