Probar las siguientes leyes de los exponentes.

Si \(a\) y \(b\) son números reales, \(a\neq0\) y \(b\neq0\), y \(m\) y \(n\) enteros, entonces:

1. \(a^{n}a^{m}=a^{n+m}\)
2. \(\left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}}\)
3. \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^{n}\)

Solución:

Ley 1.

• Caso 1. \(n>0\) y \(m>0\):
  \[ a^{n}\cdot a^{m}=\underset{n \text{ factores}}{\underbrace{a\cdot a\cdots a}}\,\underset{m \text{ factores}}{\underbrace{a\cdot a\cdots a}}=\underset{n+m \text{ factores}}{\underbrace{a\cdot a\cdots a}}=a^{n+m} \]

  \(\begin{aligned}a^{n}\cdot a^{m} &=\underset{n \text{ factores}}{\underbrace{a\cdot a\cdots a}}\,\underset{m \text{ factores}}{\underbrace{a\cdot a\cdots a}}\\[1em] &=\underset{n+m \text{ factores}}{\underbrace{a\cdot a\cdots a}}\\[1em] &=a^{n+m} \end{aligned} \)

• Caso 2. \(n<0\) y \(m<0\): Sea \(j=-n\) y \(k=-m\), entonces \(j>0\) y \(k>0\).

  \(\begin{aligned} a^{n}\cdot a^{m} =a^{-j}a^{-k}&=\dfrac{1}{a^{j}}\dfrac{1}{a^{k}}=\dfrac{1}{a^{j}a^{k}}=\dfrac{1}{a^{j+k}}=a^{-\left(j+k\right)}\\[1em] &=a^{-j-k}=a^{n+m} \end{aligned} \)

  \(\begin{aligned} a^{n}\cdot a^{m} &=a^{-j}a^{-k}\\[1em] &=\dfrac{1}{a^{j}}\dfrac{1}{a^{k}}\\[1em] &=\dfrac{1}{a^{j}a^{k}}=\dfrac{1}{a^{j+k}}\\[1em] &=a^{-\left(j+k\right)}\\[1em] &=a^{-j-k}=a^{n+m} \end{aligned} \)

• Caso 3. \(n=0\): \[ a^{n}a^{m}=a^{0}a^{m}=1\cdot a^{m}=a^{m}=a^{0+m}=a^{n+m} \]

En los otros casos, se procede en forma similar.

Ley 2.

• Caso 1. \(n>0\): \[ \dfrac{a^{n}}{b^{n}}=\dfrac{\overset{n \text{ factores}}{\overbrace{a\cdot a\cdots a}}}{\underset{n \text{ factores}}{\underbrace{b\cdot b\cdots b}}}=\underset{n \text{ factores}}{\underbrace{\dfrac{a}{b}\cdot\dfrac{a}{b}\cdots\dfrac{a}{b}}}=\left(\dfrac{a}{b}\right)^{n} \]
• Caso 2. \(n<0\):
  \[ \left(\dfrac{a}{b}\right)^{n} = \left( \dfrac{a}{b}\right)^{-k} = \dfrac{1}{\left(\dfrac{a}{b}\right)^{k}}=\dfrac{1}{\dfrac{a^{k}}{b^{k}}}=\dfrac{b^{k}}{a^{k}}=\dfrac{a^{-k}}{b^{-k}}=\dfrac{a^{n}}{b^{n}} \]

  \( \begin{aligned} \left(\dfrac{a}{b}\right)^{n} &= \left( \dfrac{a}{b}\right)^{-k}\\[1em] &= \dfrac{1}{\left(\dfrac{a}{b}\right)^{k}}\\[1em] &=\dfrac{1}{\dfrac{a^{k}}{b^{k}}}\\[1em] &=\dfrac{b^{k}}{a^{k}}\\[1em] &=\dfrac{a^{-k}}{b^{-k}}\\[1em] &=\dfrac{a^{n}}{b^{n}} \end{aligned} \)

• Caso 3. \(n=0\):

  \(\left(\dfrac{a}{b}\right)^{n}=\left(\dfrac{a}{b}\right)^{0}=1 \)   y   \( \dfrac{a^{n}}{b^{n}}=\dfrac{a^{0}}{b^{0}}=\dfrac{1}{1}=1 \)

  En consecuencia,

  \[ \left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}} \]

Ley 3.

• Caso 1. \(n>0\): \[ \left(\dfrac{a}{b}\right)^{-n}=\dfrac{1}{\left(\dfrac{a}{b}\right)^{n}}=\dfrac{1}{\dfrac{a^{n}}{b^{n}}}=\dfrac{b^{n}}{a^{n}}=\left(\dfrac{b}{a}\right)^{n} \]
• Caso 2. \(n<0:\) Si \(k=-n,\) entonces \(k>0\) y \(n=-k\).

  \(\begin{aligned} \left(\dfrac{a}{b}\right)^{-n} &=\left(\dfrac{a}{b}\right)^{k}=\dfrac{a^{k}}{b^{k}}=a^{k}\dfrac{1}{b^{k}}=a^{-\left(-k\right)}\dfrac{1}{b^{-\left(-k\right)}}=\dfrac{1}{a^{-k}}b^{-k}\\[1em] &=\dfrac{1}{a^{n}}b^{n}=\dfrac{b^{n}}{a^{n}}=\left(\dfrac{b}{a}\right)^{n} \end{aligned} \)

  \(\begin{aligned} \left(\dfrac{a}{b}\right)^{-n} &=\left(\dfrac{a}{b}\right)^{k}=\dfrac{a^{k}}{b^{k}}\\[1em] &=a^{k}\dfrac{1}{b^{k}}\\[1em] &=a^{-\left(-k\right)}\dfrac{1}{b^{-\left(-k\right)}}\\[1em] &=\dfrac{1}{a^{-k}}b^{-k}\\[1em] &=\dfrac{1}{a^{n}}b^{n}\\[1em] &=\dfrac{b^{n}}{a^{n}}\\[1em] &=\left(\dfrac{b}{a}\right)^{n} \end{aligned} \)