5.8: Nguvu ya utawala wa quotient kwa wafuasi
- Page ID
- 164586
Nguvu ya utawala wa quotient kwa watazingatia kile kinachotokea kwa quotient wakati inafufuliwa kwa nguvu fulani.
Kwa idadi yoyote halisi\(a\) na\(b\) na integer yoyote\(n\), nguvu ya utawala quotient kwa exponents ni yafuatayo:
\(\left( \dfrac{a }{b} \right)^n = \dfrac{a^n }{b^n }\),
wapi\(b \neq 0\).
Kurahisisha zifuatazo kwa kutumia nguvu ya utawala quotient kwa exponents.
Kurahisisha zifuatazo kwa kutumia nguvu ya utawala quotient kwa exponents.
\(\left( \dfrac{a }{b} \right)^4\)
Suluhisho
\(\begin{aligned} &\left( \dfrac{a}{ b} \right)^4 && \text{Given} \\ &= \dfrac{a }{b} \cdot \dfrac{a }{b} \cdot \dfrac{a }{b} \cdot \dfrac{a }{b} &&\text{Expand using the exponent definition} \\ &= \dfrac{a^4 }{b^4} && \text{Multiply as needed to simplify} \end{aligned}\)
\(\left(\dfrac{x^2 }{3y^5} \right)^3\)
Suluhisho
\(\begin{aligned} &\left( \dfrac{x^2 }{3y^5 }\right)^3 && \text{Given} \\ &= \dfrac{x^{2\cdot 3 }}{3^3 \cdot y^{5\cdot 3 }} && \text{power of quotient rule for exponents applied} \\ &= \dfrac{x^6 }{3^3 \cdot y^{15 }} &&\text{Simplify exponent product} \\ &= \dfrac{x^6 }{27y^{15 }} && \text{Multiply as needed to simplify.} \end{aligned}\)
\(\left( \dfrac{2x }{y }\right)^{−3}\)
Suluhisho
\(\begin{aligned} &\left( \dfrac{2x }{y }\right)^{−3 } &&\text{Given} \\ &= \left( \dfrac{y }{2x} \right)^3 && \text{Negative exponent rule applied} \\ &= \dfrac{y^3 }{2^3 \cdot x^3} && \text{Power of a quotient rule for exponents applied.} \\ &= \dfrac{y^3 }{8x^3 } && \text{Multiply as needed to simplify.} \end{aligned}\)
Utaratibu ambao sheria za watazamaji hutumiwa haijalishi. Kwa mfano tatu, hatua 2 na 3 zinaweza kufanywa kwa utaratibu wowote. Matokeo yatakuwa sawa.
Kurahisisha usemi kwa kutumia nguvu ya utawala quotient kwa exponents.
- \(\left( \dfrac{p^4 }{p^7 }\right) ^3\)
- \(−\left(\dfrac{ x^2 \cdot x^3 }{x \cdot y^3} \right) ^2\)
- \(\left( \dfrac{5x^3 }{2y^{13 }}\right) ^{−2}\)
- \(\left( \dfrac{2c^3}{ c^4} \right) ^3\)
- \(\left( \dfrac{a ^{−7}b }{a^2b^{−4 }}\right)^3\)
- \(\left( \dfrac{f^{−7 }}{f^5 }\right)^9\)
- \(\left(\dfrac{ xy^2z^3}{ x^3y^2z} \right) ^5\)