$$\begin{array}{ll}{a}^m{a}^n=a^{m+n}& \text{(1)}\\ \\ (a^m{)}^n=a^{mn}& \text{(2)}\\ \\ (ab{)}^n=a^n{b}^n& \text{(3)}\end{array}$$

$$\begin{array}{ll}{\log }_b(xy)={\log }_{b}x+{\log }_{b}y& \text{(1)}\\ \\ {\log }_b(x^k)=k{\log }_{b}x& \text{(2)}\\ \\ {\log }_{b}x={\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}& \text{(3)}\end{array}$$

Let $a$ and $b$ be real numbers, and let $m$ and $n$ be positive integers.

Definition (Exponentiation):

$$\begin{array}{ll}{a}^1=a,& \text{(Def. 1)}\\ a^{n+1}=a^{n}a\text{for }n\ge 1.& \text{(Def. 2)}\end{array}$$

By induction on $n$:

(1) $a^m{a}^n=a^{m+n}$

Base case ($n=1$):

$$\begin{array}{rlrl}{a}^m{a}^1& =a^{m}a& & \text{(by Def. 1)}\\ & =a^{m+1}.& & \text{(by Def. 2)}\end{array}$$

Inductive step: Assume $a^m{a}^n=a^{m+n}$. Then:

$$\begin{array}{rlrl}{a}^m{a}^{n+1}& =a^m(a^{n}a)& & \text{(by Def. 2)}\\ & =(a^m{a}^n)a& & \text{(by associativity)}\\ & =a^{m+n}a& & \text{(by the induction hypothesis)}\\ & =a^{(m+n)+1}& & \text{(by Def. 2)}\\ & =a^{m+(n+1)}.& & \text{(by associativity)}\end{array}$$

(2) $(a^m{)}^n=a^{mn}$

Base case ($n=1$):

$$\begin{array}{rlrl}(a^m{)}^1& =a^m& & \text{(by Def. 1)}\\ & =a^{m\cdot 1}.& & \text{(by identity)}\end{array}$$

Inductive step: Assume $(a^m{)}^n=a^{mn}$. Then:

$$\begin{array}{rlrl}(a^m{)}^{n+1}& =(a^m{)}^n{a}^m& & \text{(by Def. 2)}\\ & =a^{mn}{a}^m& & \text{(by the induction hypothesis)}\\ & =a^{mn+m}& & \text{(by Rule 1)}\\ & =a^{m(n+1)}.& & \text{(by distributivity)}\end{array}$$

(3) $(ab{)}^n=a^n{b}^n$

Base case ($n=1$):

$$\begin{array}{rlrl}(ab{)}^1& =ab& & \text{(by Def. 1)}\\ & =a^1{b}^1.& & \text{(by Def. 1)}\end{array}$$

Inductive step: Assume $(ab{)}^n=a^n{b}^n$. Then:

$$\begin{array}{rlrl}(ab{)}^{n+1}& =(ab{)}^n(ab)& & \text{(by Def. 2)}\\ & =a^n{b}^n(ab)& & \text{(by the induction hypothesis)}\\ & =(a^{n}a)(b^{n}b)& & \text{(by associativity and commutativity)}\\ & =a^{n+1}{b}^{n+1}.& & \text{(by Def. 2)}\end{array}$$

Let $x,y>0$ and $k$ be real numbers, and let $b,c>0$ be real numbers with $b,c\ne 1$. Since a logarithm is the inverse of exponentiation, we have:

$$b^{{\log }_{b}x}={\log }_b(b^x)=x,$$

just like $f(f^{-1}(x))=f^{-1}(f(x))=x$. Using this:

(1)

$$\begin{array}{rl}{\log }_b(xy)& ={\log }_b(b^{{\log }_{b}x}{b}^{{\log }_{b}y})\\ & ={\log }_b(b^{{\log }_{b}x+{\log }_{b}y})\\ & ={\log }_{b}x+{\log }_{b}y.\end{array}$$

(2)

$$\begin{array}{rl}{\log }_b(x^k)& ={\log }_b((b^{{\log }_{b}x}{)}^k)\\ & ={\log }_b(b^{k{\log }_{b}x})\\ & =k{\log }_{b}x.\end{array}$$

(3)

$$\begin{array}{rl}{\log }_{b}x& ={\log }_{b}x\cdot \frac{{\log }_{c}b}{{\log }_{c}b}\\ & =\frac{{\log }_c(b^{{\log }_{b}x})}{{\log }_{c}b}\\ & =\frac{{\log }_{c}x}{{\log }_{c}b}.\end{array}$$