$a × a × a × a …$ n times = $a^n$

$a$ is called the

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)

$A^m × A^n = A^(m + n)$

$A^m / A^n = A^(m - n)$

$(A^m )^n = A^{m*n}$

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. $√4=±2$

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.

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)

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