[tex]a = 8^{-2} \ km = (2^{3})^{-2} \ km = 2^{3\cdot(-2)} \ km = 2^{-6} \ km\\\\b = 8^{-3} \ km = (2^{3})^{-3} \ km = 2^{3\cdot(-3)} \ km = 2^{-9} \ km\\\\h = a = 2^{-6} \ km[/tex]
[tex]P = ?[/tex]
Pole trapezu obliczamy wg wzoru:
[tex]P = \frac{a+b}{2}\cdot h\\\\P = \frac{2^{-6}+2^{-9}}{2}\cdot2^{-6}\\\\P = \frac{2^{-6}+2^{-6}\cdot2^{-3}}{2}\cdot2^{-6}\\\\P = \frac{2^{-6}(1+2^{-3})}{2}\cdot2^{-6}\\\\P = \frac{2^{-6}\cdot2^{-6}(1+2^{-3})}{2}\\\\P = \frac{2^{-12}}{2}\cdot(1+\frac{1}{2^{3}})\\\\P = 2^{-12-1}\cdot(\frac{2^{3}+1}{2^{3}})}\\\\P = 2^{-13}\cdot(\frac{2^{3}+1}{2^{3}})\\\\P=2^{-13-3}\cdot(8+1)\\\\P = 2^{-16}\cdot 9\\\\\boxed{P = 9\cdot2^{-16} \ km^{2}}[/tex]
Wykorzystano prawa potęgowania
[tex]\\\\a^{1} = a\\\\a^{n}\cdot a^{m} = a^{n+m}\\\\a^{n}:a^{m} = a^{n-m}\\\\(a^{n})^{m} = a^{n\cdot m}\\\\a^{-n} = \frac{1}{a^{n}} \ \ \ (dla \ a\neq 0)[/tex]
