第1章前提知識Prerequisites

Simplifying Exponential Expressions指数の式を簡単にする

Recall that to simplify an expression means to rewrite it by combining terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.

英語のヒント

式を簡単にするとは、項や指数をまとめて書き直すこと、つまり項を減らしてより単純に書くことだったのを思い出そう。指数の規則を組み合わせれば、式を簡単にできる。

Example 9例9

Simplifying Exponential Expressions指数の式を簡単にする

Simplify each expression and write the answer with positive exponents only.

各式を簡単にし、答えは正の指数だけで書け。

ⓐ ${(6 m^{2} n^{−1})}^{3}$ ⓑ $17^{5} \cdot 17^{−4} \cdot 17^{−3}$ ⓒ ${(\frac{u^{−1} v}{v^{−1}})}^{2}$ ⓓ $(−2 a^{3} b^{−1}) (5 a^{−2} b^{2})$ ⓔ ${(x^{2} \sqrt{2})}^{4} {(x^{2} \sqrt{2})}^{−4}$ ⓕ $\frac{{(3 w^{2})}^{5}}{{(6 w^{−2})}^{2}}$

Solution 解答

ⓐ
$\begin{matrix} {(6 m^{2} n^{−1})}^{3} & = & {(6)}^{3} {(m^{2})}^{3} {(n^{−1})}^{3} & The power of a product rule \\ = & 6^{3} m^{2 \cdot 3} n^{−1 \cdot 3} & The power rule \\ = & 216 m^{6} n^{−3} & Simplify . \\ = & \frac{216 m^{6}}{n^{3}} & The negative exponent rule \end{matrix}$ ⓑ
$\begin{matrix} 17^{5} \cdot 17^{−4} \cdot 17^{−3} & = & 17^{5 - 4 - 3} & The product rule \\ = & 17^{−2} & Simplify . \\ = & \frac{1}{17^{2}} or \frac{1}{289} & The negative exponent rule \end{matrix}$ ⓒ
$\begin{matrix} {(\frac{u^{−1} v}{v^{−1}})}^{2} & = & \frac{{(u^{−1} v)}^{2}}{{(v^{−1})}^{2}} & The power of a quotient rule \\ = & \frac{u^{−2} v^{2}}{v^{−2}} & The power of a product rule \\ = & u^{−2} v^{2 - (−2)} & The quotient rule \\ = & u^{−2} v^{4} & Simplify . \\ = & \frac{v^{4}}{u^{2}} & The negative exponent rule \end{matrix}$ ⓓ
$\begin{matrix} (−2 a^{3} b^{- 1}) (5 a^{−2} b^{2}) & = & −2 \cdot 5 \cdot a^{3} \cdot a^{−2} \cdot b^{−1} \cdot b^{2} & Commutative and associative laws of multiplication \\ = & −10 \cdot a^{3 - 2} \cdot b^{−1 + 2} & The product rule \\ = & −10 a b & Simplify . \end{matrix}$ ⓔ
$\begin{matrix} {(x^{2} \sqrt{2})}^{4} {(x^{2} \sqrt{2})}^{−4} & = & {(x^{2} \sqrt{2})}^{4 - 4} & The product rule \\ = & {(x^{2} \sqrt{2})}^{0} & Simplify . \\ = & 1 & The zero exponent rule \end{matrix}$ ⓕ
$\begin{matrix} \frac{{(3 w^{2})}^{5}}{{(6 w^{−2})}^{2}} & = & \frac{{(3)}^{5} \cdot {(w^{2})}^{5}}{{(6)}^{2} \cdot {(w^{−2})}^{2}} & The power of a product rule \\ = & \frac{3^{5} w^{2 \cdot 5}}{6^{2} w^{−2 \cdot 2}} & The power rule \\ = & \frac{243 w^{10}}{36 w^{−4}} & Simplify . \\ = & \frac{27 w^{10 - (−4)}}{4} & The quotient rule and reduce fraction \\ = & \frac{27 w^{14}}{4} & Simplify . \end{matrix}$

式中の英語：The power of a product rule：積の累乗の法則／ The power rule：累乗の指数法則／ Simplify：整理する／ The negative exponent rule：負の指数の法則／ The product rule：積の指数法則／ or：または

Try It #9やってみよう9

Simplify each expression and write the answer with positive exponents only.

各式を簡単にし、答えは正の指数だけで書け。

ⓐ ${(2 u v^{- 2})}^{−3}$ ⓑ $x^{8} \cdot x^{−12} \cdot x$ ⓒ ${(\frac{e^{2} f^{- 3}}{f^{−1}})}^{2}$ ⓓ $(9 r^{−5} s^{3}) (3 r^{6} s^{−4})$ ⓔ ${(\frac{4}{9} t w^{−2})}^{−3} {(\frac{4}{9} t w^{−2})}^{3}$ ⓕ $\frac{{(2 h^{2} k)}^{4}}{{(7 h^{−1} k^{2})}^{2}}$