Key to "Use the rule of exponents to perform the..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Irvin Dukes

Answered question

2021-12-26

Use the rule of exponents to perform the operations and to simplify each of the following, if possible.
a)

{(- 3 x)}^{2} =

5^{- \frac{2}{3}} =

{(10^{0} - 2)}^{2} =

y^{4} \cdot x^{4} =

10^{- 5} \cdot 10^{2} =

\frac{y^{- 5}}{y^{- 2}} =

\frac{6^{10}}{6^{9}} =

{(\frac{2}{2 a})}^{- 4} =

{(2^{- 3})}^{5} =

{(- 3 x)}^{- 2} =

Answer & Explanation

Mary Nicholson

Beginner2021-12-27Added 38 answers

{(- 3 x)}^{2} = {(- 3)}^{2} x^{2} = 9 x^{2}

5^{- \frac{2}{3}} = \frac{1}{5^{\frac{2}{3}}} = \frac{1}{\sqrt[3]{5^{2}}} = \frac{1}{\sqrt[3]{25}}

{(10^{0} - 2)}^{2} = {(1 - 2)}^{2} = {(- 1)}^{2} = 1

Ethan Sanders

Beginner2021-12-28Added 35 answers

y^{4} \cdot x^{4} = {(y x)}^{4}

10^{- 5} \cdot 10^{2} = 10^{- 5 + 2} = 10^{- 3} = \frac{1}{10^{3}} = \frac{1}{1000}

\frac{y^{- 5}}{y^{- 2}} = y^{- 5 - (- 2)} = y^{- 5 + 2} = y^{- 3} = \frac{1}{y^{3}}

\frac{6^{10}}{6^{9}} = 6^{10 - 9} = 6

karton

Expert2021-12-30Added 613 answers

h) $(\frac{2}{2 a})^{- 4} = (\frac{1}{a})^{- 4} = a^{4}$
i) $(2^{- 3})^{5} = 2^{- 3 * 5} = 2^{- 15} = \frac{1}{2^{15}} = \frac{1}{32768}$
j) $(- 3 x)^{- 2} = \frac{1}{(- 3 x)^{2}} = \frac{1}{(- 3)^{2} x^{2}} = \frac{1}{9 x^{2}}$

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