第1章前提知識Prerequisites

Finding the Power of a Quotient商の累乗を求める

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.

英語のヒント

二つの式の商の累乗を簡単にするには、商の累乗の規則を使う。これは、因数の商の累乗が各因数の累乗の商になる、という規則である。たとえば次の例を見よう。

{(e^{−2} f^{2})}^{7} = \frac{f^{14}}{e^{14}}

Let’s rewrite the original problem differently and look at the result.

英語のヒント

もとの問題を別の形に書き直して、その結果を見よう。

\begin{matrix} {(e^{−2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

英語のヒント

最後の二段階から、積の累乗の規則を商の累乗の規則としても使えることが見てとれる。

\begin{matrix} {(e^{- 2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{{(f^{2})}^{7}}{{(e^{2})}^{7}} \\ = & \frac{f^{2 \cdot 7}}{e^{2 \cdot 7}} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

The Power of a Quotient Rule of Exponents 商の累乗の法則

For any real numbers $a$ and $b$ and any integer $n,$ the power of a quotient rule of exponents states that

英語のヒント

どの実数 $a$ 、 $b$ と、どの整数 $n$ についても、指数の商の累乗の規則は次を述べる。

{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}

Example 8例8

Using the Power of a Quotient Rule商の累乗の規則を使う

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

商の累乗の規則を使って、次の各商をできるだけ簡単にせよ。答えは正の指数で書け。

ⓐ ${(\frac{4}{z^{11}})}^{3}$ ⓑ ${(\frac{p}{q^{3}})}^{6}$ ⓒ ${(\frac{−1}{t^{2}})}^{27}$ ⓓ ${(j^{3} k^{−2})}^{4}$ ⓔ ${(m^{−2} n^{−2})}^{3}$

Solution 解答

ⓐ ${(\frac{4}{z^{11}})}^{3} = \frac{{(4)}^{3}}{{(z^{11})}^{3}} = \frac{64}{z^{11 \cdot 3}} = \frac{64}{z^{33}}$ ⓑ ${(\frac{p}{q^{3}})}^{6} = \frac{{(p)}^{6}}{{(q^{3})}^{6}} = \frac{p^{1 \cdot 6}}{q^{3 \cdot 6}} = \frac{p^{6}}{q^{18}}$ ⓒ ${(\frac{−1}{t^{2}})}^{27} = \frac{{(−1)}^{27}}{{(t^{2})}^{27}} = \frac{−1}{t^{2 \cdot 27}} = \frac{−1}{t^{54}} = - \frac{1}{t^{54}}$ ⓓ ${(j^{3} k^{−2})}^{4} = {(\frac{j^{3}}{k^{2}})}^{4} = \frac{{(j^{3})}^{4}}{{(k^{2})}^{4}} = \frac{j^{3 \cdot 4}}{k^{2 \cdot 4}} = \frac{j^{12}}{k^{8}}$ ⓔ ${(m^{−2} n^{−2})}^{3} = {(\frac{1}{m^{2} n^{2}})}^{3} = \frac{{(1)}^{3}}{{(m^{2} n^{2})}^{3}} = \frac{1}{{(m^{2})}^{3} {(n^{2})}^{3}} = \frac{1}{m^{2 \cdot 3} \cdot n^{2 \cdot 3}} = \frac{1}{m^{6} n^{6}}$

Try It #8やってみよう8

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

英語のヒント

商の累乗の規則を使って、次の各商をできるだけ簡単にせよ。答えは正の指数で書け。

ⓐ ${(\frac{b^{5}}{c})}^{3}$ ⓑ ${(\frac{5}{u^{8}})}^{4}$ ⓒ ${(\frac{−1}{w^{3}})}^{35}$ ⓓ ${(p^{−4} q^{3})}^{8}$ ⓔ ${(c^{−5} d^{−3})}^{4}$