The power rules

We summarise the various laws or rules you will need to calculate with powers.

Theorem 1

The following rules are valid:

Dau=a⋅...⋅a⏟udefinition a−u=1aua0=1Za0=1zero-lawSBam⋅an=am+nsame base lawaman=am−nSEam⋅bm=(a⋅b)msame exponent lawambm=(ab)mPP(am)n=am⋅npower of a power\begin{array}{r|l|l}\hline D & a^u = \underbrace{a\cdot ... \cdot a}_{u} &\text{definition }\\ & a^{-u} = \frac{1}{a^u} & \\ & a^{0} = 1 & \\\hline Z & a^0=1 &\text{zero-law}\\\hline SB & a^m \cdot a^n = a^{m+n} & \text{same base law}\\ & \frac{a^m}{a^n} = a^{m-n} & \\\hline SE & a^m \cdot b^m = (a\cdot b)^m & \text{same exponent law}\\ & \frac{a^m}{b^m}=\left(\frac{a}{b} \right)^m &\\\hline PP & \left(a^m\right)^n = a^{m\cdot n} & \text{power of a power}\\\hline \end{array}

Here, the exponents nn and mm are integers, and exponent uu is a natural number. The bases aa and bb are real numbers.

Exercise 1

Are the equations correct? Argue using the power rules and indicate where you use them.

  1. am⋅a−m=1a^m \cdot a^{-m}=1

  2. (ab)−1=ba\left(\frac{a}{b}\right)^{-1}=\frac{b}{a}

  3. 1a−m=am\frac{1}{a^{-m}}=a^m

  4. abm=(ab)mab^m = (ab)^m

  5. (1a)m=1am\left(\frac{1}{a}\right)^m=\frac{1}{a^m}

  6. (ab)−m=bmam\left(\frac{a}{b}\right)^{-m}=\frac{b^m}{a^m}

  7. (a+b)m=am+bm(a+b)^m=a^m+b^m

  8. (am)n=(an)m\left(a^m\right)^n = \left(a^n\right)^m

  9. (aaaa)m=amamamam(aaaa)^m = a^m a^m a^m a^m

Solution
  1. am⋅a−m=SBam+(−m)=a0=Z1a^m \cdot a^{-m} \overset{SB}{=} a^{m+(-m)}=a^0 \overset{Z}{=} 1 correct
  2. (ab)−1=D1(ab)1=11ab=11⋅ba=ba\left(\frac{a}{b}\right)^{-1} \overset{D}{=} \frac{1}{\left(\frac{a}{b}\right)^{1}} = \frac{\frac{1}{1}}{\frac{a}{b}} = \frac{1}{1} \cdot \frac{b}{a} = \frac{b}{a} correct
  3. 1a−m=D11am=111am=11⋅am1=am\frac{1}{a^{-m}} \overset{D}{=} \frac{1}{\frac{1}{a^m}} = \frac{\frac{1}{1}}{\frac{1}{a^m}} = \frac{1}{1} \cdot \frac{a^m}{1}= a^m correct
  4. abm=ab...b⏟m≠(ab)mab^m = a\underbrace{b ... b}_{m} \neq (ab)^m wrong
  5. (1a)m=SE1mam1am\left(\frac{1}{a}\right)^m \overset{SE}{=} \frac{1^m}{a^m} \frac{1}{a^m} correct
  6. (ab)−m=SEa−mb−m=1am1bm=1am⋅bm1=bmam\left(\frac{a}{b}\right)^{-m} \overset{SE}{=} \frac{a^{-m}}{b^{-m}} = \frac{\frac{1}{a^m} }{\frac{1}{b^m} } = \frac{1}{a^m} \cdot \frac{b^m}{1} = \frac{b^m}{a^m} correct
  7. (a+b)m=D(a+b)(a+b)...(a+b)⏟m≠am+bm(a+b)^m \overset{D}{=} \underbrace{(a+b)(a+b)...(a+b)}_{m} \neq a^m+b^m wrong
  8. (am)n=PPam⋅nan⋅m=PP(an)m\left(a^m\right)^n \overset{PP}{=} a^{m\cdot n} a^{n\cdot m} \overset{PP}{=} \left(a^n\right)^m correct
  9. (aaaa)m=((aa)⋅(aa))m=SE(aa)m⋅(aa)m=SEamamamam(aaaa)^m =((aa)\cdot(aa))^m \overset{SE}{=} (aa)^m \cdot (aa)^m \overset{SE}{=} a^m a^m a^m a^m correct OR (aaaa)m=D(a4)m=PPa4m=SE(am)4=Damamamam(aaaa)^m \overset{D}{=} (a^4)^m \overset{PP}{=} a^{4m} \overset{SE}{=} (a^m)^4 \overset{D}{=} a^m a^m a^m a^m correct