Exercises: Power Laws - learn with Serlo!

Simplify the following expressions with integer exponents as far as possible.

$\left(z^{2k-5}:z^3\right)\;:\;z^k$

$90\cdot3^{n-2}-3^n$

$\left[\left(\frac x4\right)^3\right]^5:\;\left(\frac x2\right)^6$ for $x\neq 0$

$\dfrac{\left(3a-1\right)^{2k-1}}{\left(1-3a\right)^{2k+1}}$ for $a\neq \dfrac{1}{3}$

$\left(\dfrac{6\mathrm a^2\mathrm b^{-2}}{\mathrm c^{\mathrm n+1}\mathrm d^{2\mathrm n}}\right)^3:\;\left[\dfrac{2\left(\mathrm{cd}\right)^\mathrm n}{\mathrm{ab}^{-1}}\;\cdot\;\dfrac{\mathrm c^\mathrm n\mathrm d^{2\mathrm n}}{3\mathrm{ab}^{-2}}\right]^{-2}$ for $a,b,c,d \neq 0$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Apply the power laws.

$\displaystyle \left(\frac{6\mathrm a^2\mathrm b^{-2}}{\mathrm c^{\mathrm n+1}\mathrm d^{2\mathrm n}}\right)^3:\;\left[\frac{2\left(\mathrm{cd}\right)^\mathrm n}{\mathrm{ab}^{-1}}\;\cdot\;\frac{\mathrm c^\mathrm n\mathrm d^{2\mathrm n}}{3\mathrm{ab}^{-2}}\right]^{-2}$	$=$	$\displaystyle \left(\frac{6^3a^6b^{-6}}{c^{3n+3}d^{6n}}\right):\left(\frac{\mathrm{ab}^{-1}\cdot3\mathrm{ab}^{-2}}{2\left(\mathrm{cd}\right)^n\cdot c^nd^{2n}}\right)^2$
	$=$	$\displaystyle \left(\frac{6^3\mathrm a^6}{\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}\cdot\mathrm b^6}\right):\left(\frac{\mathrm a^2\mathrm b^{-2}\cdot3^2\mathrm a^2\mathrm b^{-4}}{2^2\left(\mathrm{cd}\right)^{2\mathrm n}\cdot\mathrm c^{2\mathrm n}\mathrm d^{4\mathrm n}}\right)$
	↓	Convert division into multiplication by "flipping" the fraction.
	$=$	$\displaystyle \left(\frac{6^3\mathrm a^6}{\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}\cdot\mathrm b^6}\right)\cdot\left(\frac{2^2\left(\mathrm{cd}\right)^{2\mathrm n}\cdot\mathrm c^{2\mathrm n}\mathrm d^{4\mathrm n}}{\mathrm a^2\mathrm b^{-2}\cdot3^2\mathrm a^2\mathrm b^{-4}}\right)$
	↓	Conclude.
	$=$	$\displaystyle \frac{216\mathrm a^64\mathrm c^{4\mathrm n}\mathrm d^{6\mathrm n}}{9\mathrm a^4\mathrm c^{3\mathrm n+3}\mathrm d^{6\mathrm n}}\;$
	$=$	$\displaystyle \frac{216\mathrm a^24\mathrm c^{\mathrm n-3}}9$
	$=$	$\displaystyle 96a^2c^{n-3}$

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\displaystyle \frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left[\left(-z\right)^4\right]^{3b+3}}\;:\;\frac{x^{2a}}{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left[\left(-z\right)^3\right]^{2b-1}}

Assuming that $x,y,z\;>\;0$ , $b\in \Z$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Since $x$ is greater than $0$ , you can multiply by the reciprocal. For the value of

$A=\frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left[\left(-z\right)^4\right]^{3b+3}}\;:\;\frac{x^{2a}}{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left[\left(-z\right)^3\right]^{2b-1}}$ you then obtain

$\displaystyle A$	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left[\left(-z\right)^4\right]^{3b+3}}\;:\;\frac{x^{2a}}{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left[\left(-z\right)^3\right]^{2b-1}}$
	↓	Multiply out
	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left[\left(-z\right)^4\right]^{3b+3}}\;\cdot\;\frac{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left[\left(-z\right)^3\right]^{2b-1}}{x^{2a}}$
	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-y^3\right)^{2b+5}\cdot\;\left(-z\right)^{4\left(3b+3\right)}}\;\cdot\;\frac{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left(-z\right)^{3\left(2b-1\right)}}{x^{2a}}$
	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-y\right)^{6b+15}\cdot\;\left(-z\right)^{12b+12}}\;\cdot\;\frac{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left(-z\right)^{6b-3}}{x^{2a}}$
	↓	Factor out $(- 1)$ .
	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-1\right)^{6b+15}\cdot y^{6b+15}\cdot\left(-1\right)^{12b+12}\;\cdot z^{12b+12}}\;\cdot\;\frac{\left(\mathrm{yz}\right)^{6b+10}\cdot\;\left(-1\right)^{6b-3}\cdot z^{6b-3}}{x^{2a}}$
	↓	Decompose the $(yz)^{6b+10}$ .
	$=$	$\displaystyle \frac{x^{2a+5}}{\left(-1\right)^{6b+15}\cdot y^{6b+15}\cdot\left(-1\right)^{12b+12}\;\cdot z^{12b+12}}\;\cdot\;\frac{y^{6b+10}\cdot z^{6b+10}\cdot\;\left(-1\right)^{6b-3}\cdot z^{6b-3}}{x^{2a}}$
	$=$	$\displaystyle \frac{x^{2a+5-2a}}{\left(-1\right)^{6b+15}\cdot\left(-1\right)^{12b+12}\;}\cdot y^{6b+10-\left(6b+15\right)}\cdot z^{6b+10}\cdot\;\left(-1\right)^{6b-3}\cdot z^{6b-3-\left(12b+12\right)}$
	$=$	$\displaystyle \frac{x^{2a+5-2a}}{\left(-1\right)^{6b+15}\cdot\left(-1\right)^{12b+12}\;}\cdot y^{6b+10-6b-15}\cdot z^{6b+10}\cdot\;\left(-1\right)^{6b-3}\cdot z^{6b-3-12b-12}$
	↓	Resolve the bracket and simplify.
	$=$	$\displaystyle \frac{x^5}{\left(-1\right)^{6b+15}\cdot\left(-1\right)^{12b+12}\;}\cdot y^{-5}\cdot z^{6b+10}\cdot\;\left(-1\right)^{6b-3}\cdot z^{-6b-15}$
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{6b+10-6b-15}\cdot\;\left(-1\right)^{6b-3}}{\left(-1\right)^{6b+15}\cdot\left(-1\right)^{12b+12}\;}$
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{6b-3}}{\left(-1\right)^{6b+15}\cdot\left(-1\right)^{12b+12}\;}$
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{6b-3}}{\left(-1\right)^{6b+15+12b+12}\;}$
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{6b-3}}{\left(-1\right)^{18b+27}\;}$
	↓	The factors of $(- 1)$ can be shortened.
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{6b-3-\left(18b+27\right)}}{1\;}$
	$=$	$\displaystyle \frac{x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{-12b-30}}{1\;}$
	$=$	$\displaystyle x^5\cdot y^{-5}\cdot z^{-5}\cdot\;\left(-1\right)^{-12b-30}$
	$=$	$\displaystyle \frac{x^5}{y^5\cdot z^5}\cdot\left(-1\right)^{-12b-30}$

Since $- 12 b - 30$ is even, you get $(-1)^{-12b-30}=1$ , so the final result reads $\frac{x^5}{y^5z^5}$ , or equivalently, $\left(\dfrac{x}{yz}\right)^5$ .

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$\left(\frac{2a^{-1}b^2}{3\mathrm{ac}^{-2}}\right)^{-3}$ for $a,b,c \neq 0$

$\left(\frac uv\right)^n\cdot\;\left(\frac vu\right)^{3n+4}:\;\left(\frac{-v}u\right)^{2n+1}$ for $u,v \neq 0$

$\frac{x^5+1}{x^{m+2}}-\frac{2x^2-2}{x^m}+\frac{2-x}{x^{m-2}}$ for $x\neq 0$

$\displaystyle\left(\frac{z-3}{z+5}\right)^{2p+1}\cdot\;\left(\frac{5+z}{z-3}\right)^{p+1}:\;\left(\frac{z-3}{z+5}\right)^{4p}$ for $z \not\in \{-5;3\}$

$\left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac t2-1\right)^{-1}\right]^{-2}$ for $t \not\in \{-2;0;2\}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

Get the fraction in the rightmost round bracket on a common denominator.

$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac t2-1\right)^{-1}\right]^{-2}$	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac{t-2}2\right)^{-1}\right]^{-2}$
	↓	Apply the power laws.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac1t-\left(\frac2{t-2}\right)\right]^{-2}$
	↓	Get the square bracket on a common denominator.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{1\left(t-2\right)}{t\left(t-2\right)}-\frac{t\cdot2}{t\left(t-2\right)}\right]^{-2}$
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{t-2-2t}{t\left(t-2\right)}\right]^{-2}$
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{-t-2}{t\left(t-2\right)}\right]^{-2}$
	↓	Apply the power laws.
	$=$	$\displaystyle \left(1+\frac2t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	↓	Get the round bracket on a common denominator.
	$=$	$\displaystyle \left(\frac tt+\frac2t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	$=$	$\displaystyle \left(\frac{2+t}t\right)^2\cdot\;\left[\frac{t\left(t-2\right)}{-t-2}\right]^2$
	↓	Apply the power laws to conclude both brackets.
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot t\left(t-2\right)}{t\cdot\left(-t-2\right)}\right)^2$
	↓	Shorten by $t$ .
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot\left(t-2\right)}{\left(-t-2\right)}\right)^2$
	↓	Factor out a $(- 1)$ .
	$=$	$\displaystyle \left(\frac{\left(2+t\right)\cdot\left(t-2\right)}{-1\left(t+2\right)}\right)^2$
	↓	Shorten.
	$=$	$\displaystyle \left(\frac{t-2}{-1}\right)^2$
	$=$	$\displaystyle (-(t-2))^2$
	$=$	$\displaystyle \left(t-2\right)^2$

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Re-formulate the expression to a single fraction that does not contain any negative exponents.

$\frac{4a^{-1}z^2}{\left(x^2y\right)^3}\;:\;\frac{\left(2a\right)^{-3}}{\left(\mathrm{xy}^2z\right)^{-2}}$

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$\displaystyle \left(z^{2k-5}:z^3\right)\;:\;z^{k\;}$	$=$	$\displaystyle \left(z^{2k-5-3}\right):z^k$
	$=$	$\displaystyle z^{2k-8}:z^k$
	$=$	$\displaystyle z^{2k-8-k}$
	$=$	$\displaystyle z^{k-8}$

$\displaystyle \left[\left(\frac x4\right)^3\right]^5:\;\left(\frac x2\right)^6$	$=$	$\displaystyle \left(\frac x4\right)^{3\cdot5}:\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac x4\right)^{15}:\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right):\left(\frac{x^6}{2^6}\right)$
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right)\cdot\left(\frac{2^6}{x^6}\right)$
	↓	Write $2^{6}$ as $4^{3}$ , so you can shorten.
	$=$	$\displaystyle \left(\frac{x^{15}}{4^{15}}\right)\cdot\left(\frac{4^3}{x^6}\right)$
	$=$	$\displaystyle \frac{x^9}{4^{12}}$

$\displaystyle \frac{\left(3a-1\right)^{2k-1}}{\left(1-3a\right)^{2k+1}}$	$=$	$\displaystyle \frac{\left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1}}{\left(-3a+1\right)^{2k+1}}$
	↓	Divide and apply the power laws.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1-\left(2k+1\right)}$
	↓	Multiply out.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{2k-1-2k-1}$
	$=$	$\displaystyle \left(-1\right)^{2k-1}\left(-3a+1\right)^{-2}$
	↓	A negative exponent corresponds to a fraction.
	$=$	$\displaystyle \left(-1\right)^{2k-1}\cdot\frac1{\left(-3a+1\right)^2}$
	$=$	$\displaystyle \frac{\left(-1\right)^{2k-1}}{\left(-3a+1\right)^2}$
	$=$	$\displaystyle -\frac1{(-3a+1)^2}$

$\displaystyle \left(\frac{2a^{-1}b^2}{3\mathrm{ac}^{-2}}\right)^{-3}$	$=$	$\displaystyle \frac{2^{-3}a^3b^{-6}}{3^{-3}a^{-3}c^6}$
	$=$	$\displaystyle \frac{a^33^3a^3}{2^3c^6b^6}$
	$=$	$\displaystyle \frac{a^3\cdot27a^3}{8c^6b^6}$
	$=$	$\displaystyle \frac{27a^6}{8b^6c^6}$

$\displaystyle \displaystyle \left(\frac uv\right)^n\cdot\;\left(\frac vu\right)^{3n+4}:\;\left(\frac{-v}u\right)^{2n+1}$	$=$	$\displaystyle \displaystyle \frac{u^n}{v^n}\cdot\frac{v^{3n+4}}{u^{3n+4}}:\;\frac{-v^{2n+1}}{u^{2n+1}}$
	↓	Write the division as a fraction.
	$=$	$\displaystyle \displaystyle \frac{\frac{u^n}{v^n}\cdot\frac{v^{3n+4}}{u^{3n+4}}}{\frac{-v^{2n+1}}{u^{2n+1}}}$
	↓	Resolve the double fraction.
	$=$	$\displaystyle \displaystyle \frac{u^n}{v^n}\cdot\frac{v^{3n+4}}{u^{3n+4}}\cdot\frac{u^{2n+1}}{-v^{2n+1}}$
	$=$	$\displaystyle \displaystyle \frac{u^n\cdot v^{3n+4}\cdot u^{2n+1}}{v^n\cdot u^{3n+4}\cdot(-v^{2n+1})}$
	$=$	$\displaystyle \displaystyle \frac{u^{n+2n+1}\cdot v^{3n+4}}{u^{3n+4}\cdot\left(-1\right)\cdot v^{n+2n+1}}$
	$=$	$\displaystyle \displaystyle u^{3n+1-(3n+4)}\cdot \left(-1\right)\cdot v^{3n+4-(3n+1)}$
	↓	Shorten by using the power laws.
	$=$	$\displaystyle \displaystyle u^{-3}\cdot \left(-1\right)\cdot v^{3}$
	$=$	$\displaystyle \displaystyle \frac{\left(-1\right)\cdot v^3}{u^3}$
	$=$	$\displaystyle \displaystyle \frac{-v^3}{u^3}$
	$=$	$\displaystyle \displaystyle -\left(\frac vu\right)^3$

$\displaystyle 90\cdot3^{n-2}-3^n$	$=$	$\displaystyle 10\cdot3^2\cdot3^{n-2}-3^n$
	↓	Apply the power laws.
	$=$	$\displaystyle 10\cdot3^{2+n-2}-3^n$
	$=$	$\displaystyle 10\cdot3^n-3^n$
	↓	Factorize $3^{n}$ .
	$=$	$\displaystyle 3^n\cdot(10-1)$
	$=$	$\displaystyle 9\cdot3^n$
	↓	Write $9$ as $3^{2}$ .
	$=$	$\displaystyle 3^2\cdot3^n$
	↓	Apply the power laws.
	$=$	$\displaystyle 3^{2+n}$

$\displaystyle \frac{x^5+1}{x^{m+2}}-\frac{2x^2-2}{x^m}+\frac{2-x}{x^{m-2}}$	$=$	$\displaystyle \frac{x^5+1}{x^{m+2}}-\frac{2x^4-2x^2}{x^{m+2}}+\frac{2-x}{x^m\cdot x^{-2}}$
	↓	Apply the power laws to $x^{- 2}$ .
	$=$	$\displaystyle \frac{x^5+1}{x^{m+2}}-\frac{2x^4-2x^2}{x^{m+2}}+\frac{2x^2-x^3}{x^m}$
	↓	Get to a common denominator.
	$=$	$\displaystyle \frac{x^5+1}{x^{m+2}}-\frac{2x^4-2x^2}{x^{m+2}}+\frac{2x^4-x^5}{x^{m+2}}$
	$=$	$\displaystyle \frac{x^5+1-2x^4+2x^2+2x^4-x^5}{x^{m+2}}$
	$=$	$\displaystyle \frac{1+2x^2}{x^{m+2}}$

$\displaystyle \left(\frac{z-3}{z+5}\right)^{2p+1}\cdot\;\left(\frac{5+z}{z-3}\right)^{p+1}:\;\left(\frac{z-3}{z+5}\right)^{4p}$	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{2p+1-4p}\cdot\;\left(\frac{5+z}{z-3}\right)^{p+1}$
	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{1-2p}\cdot\;\left(\frac{5+z}{z-3}\right)^{p+1}$
	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{1-2p}\cdot\;\left(\frac{z-3}{z+5}\right)^{-\left(p+1\right)}$
	↓	Resolve the bracket.
	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{1-2p}\cdot\;\left(\frac{z-3}{z+5}\right)^{-p-1}$
	↓	Apply the power laws.
	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{1-2p+(-p)-1}$
	$=$	$\displaystyle \left(\frac{z-3}{z+5}\right)^{-3p}$
	$=$	$\displaystyle \left(\frac{z+5}{z-3}\right)^{3p}$

$\displaystyle \frac{4a^{-1}z^2}{\left(x^2y\right)^3}\;:\;\frac{\left(2a\right)^{-3}}{\left(\mathrm{xy}^2z\right)^{-2}}$	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}:\frac{\frac1{\left(2a\right)^3}}{\frac1{\left(\mathrm{xy}^2z\right)^2}}$
	↓	Division can be done by "flipping the fraction".
	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}\cdot\frac{\frac1{\left(\mathrm{xy}^2z\right)^{-2}}}{\frac1{\left(2a\right)^3}}$
	↓	Resolve the double fraction.
	$=$	$\displaystyle \frac{4\cdot\frac1a\cdot z^2}{x^6\cdot y^3}\cdot\frac1{\left(\mathrm{xy}^2z\right)^2}\cdot\frac{\left(2a\right)^3}1$
	$=$	$\displaystyle \frac{\frac{4z^2}a}{x^6\cdot y^3}\cdot\frac{8a^3}{x^2\cdot y^4\cdot z^2}$
	$=$	$\displaystyle \frac{\frac{4z^2\cdot8a^3}a}{x^{6+2}\cdot y^{3+4}\cdot z^2}$
	↓	Finally, shorten.
	$=$	$\displaystyle \frac{4\cdot8a^2}{x^8y^7}$
	$=$	$\displaystyle \frac{32a^2}{x^8y^7}$