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

(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

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

IMO B

This is how I did. Hope it helps.

Posted from my mobile device

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

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

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).

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

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

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

Answer (B)

(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.

Posted from my mobile device

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.

\((\frac{1}{3})^{−3}* (\frac{1}{9})^{−2}* (\frac{1}{27})^{−1}* (\frac{1}{81})^{0}\) =

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

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

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

The answer is B.

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


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

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