Pangkat Bulat

$\begin{aligned} 2^{-4} \cdot 2^7 &= 2^{-4+7} \\ &= 2^{3} \\ &= 8 \end{aligned}$

$\begin{aligned} (-5)^{-7} \cdot (-5)^9 &= (-5)^{-7+9} \\ &= (-5)^{2} \\ &= (-5) \cdot (-5) \\ &= 25 \end{aligned}$

$\begin{aligned} (6a^5 b^{-3}) \cdot (2a^{-2} b^7) &= 6 \cdot a^5 \cdot b^{-3} \cdot 2 \cdot a^{-2} \cdot b^7 \\ &= 6 \cdot 2 \cdot a^5 \cdot a^{-2} \cdot b^{-3} \cdot b^7 \\ &= 12 \cdot a^{5+(-2)} \cdot b^{-3+7} \\ &= 12a^3 b^4 \end{aligned}$

$\begin{aligned} (16x^5 y^3) \cdot \left( \frac{1}{2^3} x^{-3} y^{-2} \right) &= 16 \cdot x^5 \cdot y^3 \cdot \frac{1}{2^3} \cdot x^{-3} \cdot y^{-2} \\ &= 2^4 \cdot \frac{1}{2^3} \cdot x^5 \cdot x^{-3} \cdot y^3 \cdot y^{-2} \\ &= \frac{2^4}{2^3} \cdot x^{5-3} \cdot y^{3-2} \\ &= 2^{4-3} \cdot x^2 \cdot y^1 \\ &= 2x^2 y \end{aligned}$

$\begin{aligned} \left( 2 - \frac{1}{2} - \frac{1}{2^2} \right)^{-2} &= \left( 2 - \frac{1}{2} - \frac{1}{4} \right)^{-2} \\ &= \left( \frac{8}{4} - \frac{2}{4} - \frac{1}{4} \right)^{-2} \\ &= \left( \frac{8-2-1}{4} \right)^{-2} \\ &= \left( \frac{5}{4} \right)^{-2} \\ &= \frac{5^{-2}}{4^{-2}} \\ &= \frac{4^2}{5^2} \\ &= \frac{16}{25} \end{aligned}$

$\begin{aligned} a^9 : a^{-3} &= \frac{a^9}{a^{-3}} \\ &= a^{9-(-3)} \\ &= a^{9+3} \\ &= a^{12} \end{aligned}$

$\begin{aligned} \frac{3^{n+1}-3^{n-1}}{3^{n-1}-3^{n-2}} &= \frac{3^n \cdot 3^1 - \frac{3^n}{3^1}}{\frac{3^n}{3^1} - \frac{3^n}{3^2}} \\ &= \frac{3^n \left( 3 - \frac{1}{3} \right)}{3^n \left( \frac{1}{3} - \frac{1}{9} \right)} \\ &= \frac{3 - \frac{1}{3}}{\frac{1}{3} - \frac{1}{9}} \\ &= \frac{\frac{8}{3}}{\frac{2}{9}} \\ &= \frac{8}{3} \cdot \frac{9}{2} \\ &= 4 \cdot 3 \\ &= 12 \end{aligned}$

$\begin{aligned} \frac{2^{-3} a^3 b^{-4}}{\left( \frac{1}{2} \right)^4 a^4 b^{-5}} &= \frac{2^{-3} a^3 b^{-4}}{\frac{1}{2^4} a^4 b^{-5}} \\ &= \frac{2^{-3} \cdot 2^4 \cdot a^3 \cdot b^{-4}}{a^4 \cdot b^{-5}} \\ &= \frac{2^{4-3} \cdot b^{-4-(-5)}}{a^{4-3}} \\ &= \frac{2^1 \cdot b^1}{a^1} \\ &= \frac{2b}{a} \end{aligned}$

$\begin{aligned} \left( \left( \frac{1}{2} \right)^3 \right)^{-2} &= \left( \frac{1}{2} \right)^{3 \cdot (-2)} \\ &= \left( \frac{1}{2} \right)^{-6} \\ &= \frac{1^{-6}}{2^{-6}} \\ &= \frac{2^6}{1^6} \\ &= 64 \end{aligned}$

$\begin{aligned} \left( 2 \cdot (2^2 \cdot (2^3)^{-1})^{-2} \right) &= \left( 2 \cdot (2^2 \cdot 2^{-3})^{-2} \right) \\ &= \left( 2 \cdot (2^{-1})^{-2} \right) \\ &= (2 \cdot 2^2) \\ &= 2^3 \\ &= 8 \end{aligned}$

$\begin{aligned} \left( \left( \frac{1}{2} \right)^2 \cdot 2^2 \right)^{-2} &= \left( \frac{1}{4} \cdot 4 \right)^{-2} \\ &= 1^{-2} \\ &= 1 \end{aligned}$

$\begin{aligned} (2a^3 b^{-4})^2 &= 2^2 a^{3 \cdot 2} b^{-4 \cdot 2} \\ &= 4a^6 b^{-8} \\ &= \frac{4a^6}{b^8} \end{aligned}$

$\begin{aligned} \frac{(0,6)^0 - (0,1)^{-1}}{\left( \frac{3}{2^3} \right)^{-1} \left( \frac{3}{2} \right)^3 + \left( -\frac{1}{3} \right)^{-1}} &= \frac{1 - \left( \frac{1}{10} \right)^{-1}}{\left( \frac{2^3}{3} \right) \left( \frac{3^3}{2^3} \right) + (-3)} \\ &= \frac{1 - 10}{3^2 - 3} \\ &= \frac{-9}{9-3} \\ &= \frac{-9}{6} \\ &= -\frac{3}{2} \end{aligned}$

$\begin{aligned} \left( \frac{3a^6 b^{-5}}{81a^9 b^{-2}} \right)^{-1} &= \frac{81a^9 b^{-2}}{3a^6 b^{-5}} \\ &= \frac{3^4 a^9 b^{-2}}{3^1 a^6 b^{-5}} \\ &= 3^{4-1} a^{9-6} b^{-2-(-5)} \\ &= 3^3 a^3 b^3 \\ &= (3ab)^3 \end{aligned}$

$\begin{aligned} \frac{x^{-1} + y^{-1}}{x^{-1} - y^{-1}} &= \frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x} - \frac{1}{y}} \\ &= \frac{\frac{y+x}{xy}}{\frac{y-x}{xy}} \\ &= \frac{y+x}{xy} \cdot \frac{xy}{y-x} \\ &= \frac{x+y}{y-x} \end{aligned}$