$$e^{i\pi }+1=0$$

$$e^{i\theta }=\cos \theta +i\sin \theta$$

$$(a+b{)}^2=a^2+2ab+b^2$$

$$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$$

$$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$$

$$a{x}^2+bx+c=0$$

$$⟹x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$a^m{a}^n=a^{m+n}$$

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

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

$${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$$

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

$${\log }_{b}x={\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}$$

$$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$$

$$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$

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

Definition (Exponentiation):

$a^1=a,\text{(Def. 1)}$

$a^{n+1}=a^{n}a\text{for }n\ge 1.\text{(Def. 2)}$

By induction on $n$:

$a^m{a}^n=a^{m+n}$

Base case ($n=1$):

$a^m{a}^1$

$=a^{m}a\text{(by Def. 1)}$

$=a^{m+1}.\text{(by Def. 2)}$

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

$a^m{a}^{n+1}$

$=a^m(a^{n}a)\text{(by Def. 2)}$

$=(a^m{a}^n)a\text{(by associativity)}$

$=a^{m+n}a\text{(by the induction hypothesis)}$

$=a^{(m+n)+1}\text{(by Def. 2)}$

$=a^{m+(n+1)}.\text{(by associativity)}$

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

Base case ($n=1$):

$(a^m{)}^1$

$=a^m\text{(by Def. 1)}$

$=a^{m\cdot 1}.\text{(by identity)}$

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

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

$=(a^m{)}^n{a}^m\text{(by Def. 2)}$

$=a^{mn}{a}^m\text{(by the induction hypothesis)}$

$=a^{mn+m}\text{(by Rule 1)}$

$=a^{m(n+1)}.\text{(by distributivity)}$

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

Base case ($n=1$):

$(ab{)}^1$

$=ab\text{(by Def. 1)}$

$=a^1{b}^1.\text{(by Def. 1)}$

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

$(ab{)}^{n+1}$

$=(ab{)}^n(ab)\text{(by Def. 2)}$

$=a^n{b}^n(ab)\text{(by the induction hypothesis)}$

$=(a^{n}a)(b^{n}b)\text{(by associativity and commutativity)}$

$=a^{n+1}{b}^{n+1}.\text{(by Def. 2)}$

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:

${\log }_b(xy)$

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

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

$={\log }_{b}x+{\log }_{b}y.$

${\log }_b(x^k)$

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

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

$=k{\log }_{b}x.$

${\log }_{b}x$

$={\log }_{b}x\cdot {\displaystyle \frac{{\log }_{c}b}{{\log }_{c}b}}$

$={\displaystyle \frac{{\log }_c(b^{{\log }_{b}x})}{{\log }_{c}b}}$

$={\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}.$