W zadaniu korzystamy ze wzorów:

$a^m\cdot a^n=a^{m+n}$  

$\frac{b^m}{b^n}=b^m:b^n=b^{m-n}\text{dla}b\ne 0$   

 

$\text{a)}9\cdot 3^{11}=3^2\cdot 3^{11}=3^{2+11}=3^{13}$ $\text{a)}9\cdot 3^{11}=3^2\cdot 3^{11}=3^{2+11}=3^{13}$  

$\text{b)}{2}^{13}:32=2^{13}:2^5=2^{13-5}=2^8$  

$\text{c)}{2}^{50}:\left(16\cdot 2^{10}\right)=2^{50}:\left(2^4\cdot 2^{10}\right)=2^{50}:2^{14}=2^{36}$  

$\text{d)}\frac{125\cdot 5^6}{5^7}=\frac{5^3\cdot 5^6}{5^7}=\frac{5^9}{5^7}=5^2$  

$\text{e)}\frac{5^{20}}{5^5\cdot 5^{12}\cdot 25}=\frac{5^{20}}{5^5\cdot 5^{12}\cdot 5^2}=\frac{5^{20}}{5^{19}}=5^1=5$  

$\text{f)}\frac{{20}^{10}}{20\cdot {20}^6\cdot 400}=\frac{{20}^{10}}{20\cdot {20}^6\cdot {20}^2}=\frac{{20}^{10}}{2^9}={20}^1=20$  

$\text{g)}\frac{4\cdot 2^7}{8\cdot 2^5}=\frac{2^2\cdot 2^7}{2^3\cdot 2^5}=\frac{2^9}{2^8}=2^1=2$  

$\text{h)}\left(81\cdot 3^{15}\right):\left(3^7\cdot 27\right)=\left(3^4\cdot 3^{15}\right):\left(3^7\cdot 3^3\right)=3^{19}:3^{10}=3^9$