$\text{a)}{10}^{-7}>{10}^{-9}$  

${10}^{-7}=\frac{1}{{10}^7}$  

${10}^{-9}=\frac{1}{{10}^9}$  

$\frac{1}{{10}^7}>\frac{1}{{10}^9}$    - z dwóch ułamków o takich samych licznikach, ten jest większy, który ma mniejszy mianownik

$\underline{}$  

$\text{b)}{8}^{-7}>{18}^{-7}$  

$8^{-7}=\frac{1}{8^7}$   

${18}^{-7}=\frac{1}{{18}^7}$

$\frac{1}{8^7}>\frac{1}{{18}^7}$  

$\underline{}$ $\underline{}$  

`"c)"\ (1/7)^-5  

${\left(\frac{1}{7}\right)}^{-5}=7^5$

 

${\left(\frac{1}{7}\right)}^{-8}=7^8$  

$7^5<7^8$  

$\underline{}$  

$\text{d)}{\left(\frac{1}{11}\right)}^{-6}>{\left(\frac{1}{9}\right)}^{-6}$  

${\left(\frac{1}{11}\right)}^{-6}={11}^6$  

${\left(\frac{1}{9}\right)}^{-6}=9^6$  

${11}^6>9^6$  

$\underline{}$  

$\text{e)}{4}^{-9}=2^{-18}$  

$4^{-9}={\left(2^2\right)}^{-9}=2^{-18}$  

$\underline{}$  

$\text{f)}{5}^{-12}>{125}^{-5}$  

$5^{-12}=\frac{1}{5^{12}}$  

${125}^{-5}={\left(5^3\right)}^{-5}=5^{-15}=\frac{1}{5^{15}}$  

$\frac{1}{5^{12}}>\frac{1}{5^{15}}$  

$\underline{}$  

$\text{g)}{27}^{-5}<9^{-7}$  

${27}^{-5}={\left(3^3\right)}^{-5}=3^{-15}=\frac{1}{3^{15}}$   

$9^{-7}={\left(3^2\right)}^{-7}=3^{-14}=\frac{1}{3^{14}}$  

$\frac{1}{3^{15}}<\frac{1}{3^{14}}$  

$\underline{}$

$\text{h)}{16}^{-8}>8^{-12}$  

${16}^{-8}={\left(2^4\right)}^{-8}=2^{-32}=\frac{1}{2^{32}}$   

$8^{-12}={\left(2^3\right)}^{-12}=2^{-36}=\frac{1}{2^{36}}$  

$\frac{1}{2^{32}}>\frac{1}{2^{36}}$