Indices
$a × a × a × a …$ n times = $a^n$
$a$ is called the base and $n$ the exponent or index
If the index is negative, we take the reciprocal.
e.g. $4^-2 = 1/4^2 = 1/16$
If the index is a fraction, we take that root.
e.g. $4^{1/2} = √4 = ±2$
$A^0 = 1$ (any non-zero integer raised to 0 gives 1)
$A^m × A^n = A^(m + n)$
$A^m / A^n = A^(m - n)$
$(A^m )^n = A^{m*n}$

Roots
If $a × a × a$ ... n times = $x$, then the $n$th root of $x$ or $x^{1/n} =a$
$√a × √b=√(a*b); √a / √b=√(a/b)$
e.g. $2^5 =32$, so $32^(1/5) =2$ or $2$ is the $5$th root of $32$. e.g. $√8=√(2*4)=√(4)*√(2)=2√2$
e.g. please note how $(-2)*(-2)=4=2*2$. For even roots, both positive and negative roots are viable.
i.e. $√4=±2$

Logarithms
Definition of Logarithm: If $e^x = b$, then $x • \text"ln(e) = ln(b)"$ or $x = \text"ln(b)"$
Here, ln(x) is the logarithm of x to the base "e" where "e" is a transcendental number.
Other rules:
$\text"log (a • b) = log (a) + log (b)"$
$\text"log (a / b) = log (a) - log (b)"$
And the change of base theorem: $log_cx = e{\text"ln (x)"} /{\text"ln (c)"}$
Here, we changed the base from c to e, the base of natural logarithms.
e.g. Solve: $2^16 =4^{?}$ . We simplify to $2^16 =2^{2x}$ giving $x=8$ after equating the indices.

Operator Precedence Now that we have covered most arithmetic operators, we can discuss precedence.
Precedence order is: always parentheses first, then exponentiation, then multiplication and division, and finally addition and subtraction.
Operations must be performed in that order.
e.g. $2+4*5-24/4+(6+4)/2 = 2+4*5-24/4+10/2 = 2+20-24/4+10/2 = 2+20-6+10/2 = 2+20-6+5=21$.
e.g. $2^2 +3-24/6+√16=4+3-24/6+√16=4+3-24/6+4=4+3-4+4=7$.
(we considered the positive root of 16 in the above example)