$$ \underbrace {a \times a \times a \times {...} \times a \times a \times a}_{n\;tane} = a^n$$

$$ \displaystyle a \in \mathbb{R}, n \in \mathbb {Z}$$

$$ a^0 = 1$$

$$ \begin{array}{1c1l} \text{(I)} & 2^0 = (-2)^0 = \left(\sqrt{2}\right)^0 = 1 \\ \text{(II)} & \pi^0 = e^0 = \Phi^0 = 1 \\ \text{(III)} & 0^0 \: \wedge \: \infty^0 \rightarrow \unicode{x2753} \end{array}$$

$$ a^1 = a$$

$$ \begin{array}{1c1l} \text{(I)} & 2^1 = 2 \\ \text{(II)} & \pi^1 = \pi \\ \text{(III)} & 0^1 = 0 \end{array}$$

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

$$ \begin{array}{1c1l} \text{(I)} & 2^3 \times 2^4 = 2^{3 + 4} = 2^{7} \\ \text{(II)} & \pi^{20} \times \pi^{11} = \pi^{20 + 11} = \pi^{31} \end{array}$$

$$ \frac{a^m}{a^n} = a^{m - n}$$

$$ \begin{array}{1c1l} \text{(I)} & \frac{2^3}{2^4} = 2^{3 - 4} = 2^{-1} \\ \text{(II)} & \frac{\pi^{20}}{\pi^{11}} = \pi^{20 - 11} = \pi^{9} \end{array}$$

$$ a^n \times b^n = (a \times b)^{n}$$

$$ \begin{array}{1c1l} \text{(I)} & 2^2 \times 3^2 = (2 \times 3)^2 = 6^2 \\ \text{(II)} & \pi^3 \times e^3 = (\pi \times e)^3 \end{array}$$

$$ \frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$$

$$ \begin{array}{1c1l} \text{(I)} & \frac{2^2}{3^2} = \left(\frac23\right)^2 \\ \text{(II)} & \frac{\pi^3}{e^3} = \left(\frac{\pi}{e}\right)^3 \end{array}$$

$$ \left(a^m\right)^n = \left(a^n\right)^m = a^{m \times n}$$

$$ \begin{array}{1c1l} \text{(I)} & \left(2^3\right)^4 = \left(2^4\right)^3 = 2^{3 \times 4} \\ \text{(II)} & \left(\pi^{-1}\right)^{3} = \left(\pi^3\right)^{-1} = \pi^{(-1)\times 3} \end{array}$$

$$ a^{-n} = \frac{1}{a^n}$$

$$ \begin{array}{1c1l} \text{(I)} & 15^{-3} = \frac{1}{15^3} \\ \text{(II)} & \pi^{-1} = \frac{1}{\pi} \end{array}$$

$$ n \in \mathbb {Z}$$

$$ \begin{array}{1l1l} -a^m = (-a)^m & m = 2n - 1 \\ -a^m = -(-a)^m & m = 2n \\ -a^m = (-1) \times a^m & \color{green} \unicode{x2714} \\ -a^m = (-1)^m \times a^m & \color{red} \unicode{x2718} \end{array}$$

$$ \begin{array}{1c1l} \text{(I)} & -2^3 = -1 \times 2^3 = -8 \\ \text{(II)} & -\pi^{-2} = -1 \times \frac{1}{\pi^2} = -\frac{1}{\pi^2} \end{array}$$

$$ p \times a^m \pm r \times a^m = (p \pm r) \times a^m$$

$$ \begin{array}{1c1l} \text{(I)} & 2\times 3^4 - 5 \times 3^4 = (2 - 5)\times 3^4 \\ & = -3 \times 3^4 = -3^5 \\ \text{(II)} & 5 \pi^2 + 3 \pi^2 = (5 + 3)\times \pi^2 \\ & = 8 \times \pi^2 \end{array}$$

$$ \begin{array}{l11} a^m = a^n & \Rightarrow & m = n \\ a^m = b^m & \Rightarrow & a = b \end{array}$$

$$ \begin{array}{1c1l} \text{(I)} & 2^{x - 2} = 2^{1 - x} \\ &\Rightarrow x - 2 = 1 - x \\ &\Rightarrow 2x = 3 \\ &\Rightarrow x = \frac32 \\ \text{(II)} & (x - 1)^3 = (2x - 3)^3 \\ &\Rightarrow x - 1 = 2x - 3 \\ & \Rightarrow -x = -2 \\ &\Rightarrow x = 2 \end{array}$$

$$ \begin{align} \frac{3^8 - 2 \times {15}^4 + 5^8}{(9^2 - {25}^2) \times (3^2 - 5^2)} = ? \end{align}$$

$$ \begin{align} &= \frac{3^8 - 2 \times {15}^4 + 5^8}{(9^2 - {25}^2) \times (3^2 - 5^2)} \\ &= \frac{\mathtip{3^8 - 2 \times 3^4 \times 5^4 + 5^8}{(a-b)^2 = a^2 - 2ab + b^2}}{((3^2)^2 - (5^2)^2) \times (3^2 - 5^2)} \\ &= \frac{(3^4 - 5^4)^2}{(3^4 - 5^4) \times (3^2 - 5^2)} \\ &= \frac{3^4 - 5^4}{3^2 - 5^2} \\ &= \frac{(3^2 - 5^2) \times (3^2 + 5^2)}{3^2 - 5^2} = 34 \end{align}$$

$$ \begin{align} A = \frac{(25^{12}) \times (2^{26} - 2^{24})}{0,0015} \end{align}$$

$$ \begin{align} &= \frac{(25^{12}) \times (2^{26} - 2^{24})}{0,0015} \\ &= \frac{((5^2)^{12}) \times 2^{24} \times (2^2 - 1)}{1,5 \times 10^{-3}} \\ &= \frac{5^{24} \times 2^{24} \times 3}{1,5 \times 10^{-3}} \\ &= \frac{10^{24} \times 10^3 \times 3}{1,5} \\ &= 2 \times 10^{27} \end{align}$$

$$ \begin{align} A = \frac{-1^{32} \cdot ((-1)^3)^2 \cdot -2^3 \cdot (-3)^2}{-2^2 \cdot (-3)^2} \end{align}$$

$$ \begin{align} &= \frac{-1^{32} \cdot ((-1)^3)^2 \cdot -2^3 \cdot (-3)^2}{-2^2 \cdot (-3)^2} \\ &= \frac{-1 \cdot (-1)^2 \cdot -8 \cdot 9}{-4 \cdot 9} \\ &= \frac{-1 \cdot 1 \cdot -8 \cdot 9}{-4 \cdot 9} = \frac{72}{-36} = -2 \end{align}$$ $$ \begin{align} x \cdot (-2) &= 1 \Rightarrow x = -\frac12 \end{align}$$

$$ \begin{align} \frac{6^6 + 27^2}{54^2 + 3^8} = ? \end{align}$$

$$ \begin{align} &= \frac{6^6 + 27^2}{54^2 + 3^8} \\ &= \frac{2^6 \times 3^6 + \toggle{(3^3)^2}{3^6} \endtoggle}{\toggle{27^2}{3^6}\endtoggle \times 2^2 + 3^8} \\ &= \frac{3^6 \times (2^6 + 1)}{3^6 \times (2^2 + 3^2)} \\ &= \frac{\toggle{3^6}{\cancel{3^6}}\endtoggle \times 65}{\toggle{3^6}{\cancel{3^6}}\endtoggle \times 13} \\ &= 5 \end{align}$$

$$ \begin{align} \frac{3^8 - 1}{3^6 - 3^4 + 3^2 - 1} = ? \end{align}$$

$$ \begin{align} &= \frac{3^8 - 1}{3^6 - 3^4 + 3^2 - 1} \\ &= \frac{\left(3^4 - 1\right)\left(3^4 + 1\right)}{3^4 \left(3^2 - 1\right) + \left(3^2 - 1\right)} \\ &= \frac{\left(3^4 - 1\right)\left(3^4 + 1\right)}{\left(3^2 - 1\right)\left(3^4 + 1\right)} \\ &= \frac{\left(3^4 - 1\right)}{\left(3^2 - 1\right)} \\ &= \frac{\left(3^2 - 1\right)\left(3^2 + 1\right)}{\left(3^2 - 1\right)} \\ &= 3^2 + 1 = 10 \end{align}$$

$$ \left. \begin{align} 2^x - 3^y &= 9 \\ 2^{x+1} &= 3^{y+1} \end{align} \right \} \Rightarrow 2^x + 3^y = ? $$

$$ \begin{align} 2^{x + 1} &= 3^{y + 1} \\ 2 \times 2^{x} &= 3 \times 3^y \end{align}$$ $$ \begin{align} \text{(I)} \:\:\: 2 \times 2^{x} - 3 \times 3^y &= 0 \end{align}$$ $$ \begin{align} 2^x - 3^y &= 9 \\ 2 \times 2^x - 2 \times 3^y &= 2 \times 9 \end{align}$$ $$ \begin{align} \text{(II)} \:\:\: 2 \times 2^x - 2 \times 3^y &= 18 \end{align}$$ $$ \enclose{bottom} {\begin{array}{1c1l} \text{(I)} & 2 \times 2^{x} - 3 \times 3^y = 0 \\ \text{(II)} & 2 \times 2^x - 2 \times 3^y = 18 \end{array}}$$ $$ \begin{align} \text{(II - I)} \Rightarrow 3^y = 18 \end{align}$$ $$ \begin{align} 2^x - 3^y &= 9 \\ 2^x - 18 &= 9 \\ 2^x &= 27 \end{align}$$ $$ \begin{align} \Rightarrow 2^x + 3^y = 27 + 18 = 45 \end{align}$$