ZS Associates Exponents Quiz-4

\frac{(42 \times 229)}{9261^{\frac{1}{3}}} and
9621^{\frac{1}{3}} = (21 \times 21 \times 21)^{\frac{1}{3}} = 21
42 \times 229 ÷ 21 = x
42 \times 229 ÷ 21 = x
42 \times 229 \frac{1}{21} = x
2 \times 229 = x
x = 2 \times 229
x = 458

\frac{(-1)^{132}}{5^{-1}+3^{-1}}
= \frac{(-1^{2})^{66}}{5^{-1}+3^{-1}}
= \frac{(1)^{66}}{5^{-1}+3^{-1}}
= \frac{1}{\frac{1}{5}+\frac{1}{3}}
= \frac{1}{\frac{5+3}{15}} = \frac{15}{8}

(\frac{2^{3}}{3^{2}})^{\frac{2}{3}} \times \frac{3^{3}}{2^{2}}
= \frac{2^{2}}{3^{\frac{4}{3}}} \times \frac{3^{3}}{2^{2}} = \frac{3^{3}}{3^{\frac{4}{3}}} = 3^{3-4/3}=3^{\frac{5}{3}}

\frac{a^{\frac{27}{2}}}{a^{\frac{23}{2}}} \times \frac{\sqrt{b^{21}}}{\sqrt{b^{23}}}
= a^{27/2-23/2} \times b^{21/2-23/2}
= a^{2} \times b^{-1}

(\frac{3^{5}}{10^{5}})^{\frac{2}{5}} = \frac{3}{10}^{5 \times \frac{2}{5}} = \frac{3}{10}^{2} = \frac{9}{100} = 0.09

Given that x^{a} = y , y^{b}=z
Apply the law of algebra.
Take the power b of both sides of the equation.
y^{b}=x^{ab}
Take the power c of both sides of the equation.
=> y^{bc}=x^{abc}
=> (y^{b})^c=x^{ab}
Replace the y^b with z and abc with 1 as per given equation
=> z^{c}=x

5 > 4 and \frac{1}{2} > \frac{1}{4}

5^{\frac{1}{2}} > 4^{\frac{1}{4}}

Thus, 4^{\frac{1}{4}} cannot be the greatest.

3^{\frac{7}{10}} = (3^{7})^{\frac{1}{10}} = (2187)^{\frac{1}{10}}

5^{\frac{1}{2}} = 5^{\frac{5}{10}} = (5^{5})^{\frac{1}{10}} = (3125)^{\frac{1}{10}}

6^{\frac{1}{5}} = 6^{\frac{2}{10}} = (6^{2})^{\frac{1}{10}} = (36)^{\frac{1}{10}}

Thus 5^{\frac{1}{2}} is the greatest.

3^{34}=(3^{2})^{17} = 9^{17}

2^{51} = (2^{3})^{17} = 6^{17}

Thus 3^{34}>7^{17}>2^{51}

2^{122}+2^{124}+2^{126}+2^{128}

= 2^{122}(1+2^{2}+2^{4}+2^{6})=2^{122} \times 85 and 85 divisible only by 17 among all the given options

\sqrt{30+\sqrt{6\sqrt{36}}+8 \times \sqrt{8\sqrt{64}}}
=\sqrt{30+\sqrt{6 \times 6}+8 \times \sqrt{8 \times 8}} = \sqrt{30+6+64} = \sqrt{100} =10