1) Rozwiąż równanie: [latex] 2^{2x} [/latex]·16=64 2) Zapisz w postaci jednej potęgi: a) [latex] 125^{2}[/latex]·[latex] 25^{3}[/latex]·([latex] frac{1}{25})^{-2} [/latex]·[latex]{25}^{-2}[/latex] b) [latex]27:3^{2}: 9^{3}: 81^{5} [/latex]

[latex]2^{2x}cdot 16=64[/latex] [latex]2^{2x}cdot 2^4=2^6[/latex] [latex]2^{2x+4}=2^6[/latex] [latex]2x+4=6[/latex] [latex]2x=6-4[/latex] [latex]2x=2[/latex] [latex]x=1[/latex] --- [latex]125^2cdot 25^3cdot (frac{1}{25})^{-2}cdot 25^{-2}=[/latex] [latex](5^3)^2cdot (5^2)^3cdot (5^{-2})^{-2}cdot (5^2)^{-2}=[/latex] [latex]5^6cdot 5^6cdot 5^4cdot 5^{-4}=[/latex] [latex]5^{6+6+4-4}=5^{12}[/latex] --- [latex]27:3^2:9^3:81^5=3^3:3^2:(3^2)^3:(3^4)^5=\3^3:3^2:3^6:3^{20}=3^{3-2-6-20}=3^{-25}[/latex]

zad 1 [latex]2^{2x}*16=64\ \ 2^{2x}*2^{4}=2^{6}\ \ 2^{2x+4}=2^{6}\ \ 2x+4=6\ \ 2x=2 |:2\ \ x=1[/latex] zad 2 [latex]a) 125^{2}*25^{3}*(frac{1}{25})^{-2}*25^{-2} =(5^{3})^{2}*(5^{2})^{3}*(5^{-2})^{-2}*(5^{2})^{-2}=\ \ =5^{3*2}*5^{2*3}*5^{-2*(-2)}*5^{2*(-2)}= 5^{6}*5^{6}*5^{4}*5^{-4}=\ \ =5^{6+6+4-4}=5^{12}[/latex] [latex] b) 27:3^{2}:9^{3}:81^{5}= 3^{3}:3^{2}:(3^{2})^{3}:(3^{4})^{5} =3^{3-2}:3^{2*3}:3^{4*5}=\ \ =3^{1}:3^{6}:3^{20}=3^{1-6-20}=3^{-25}[/latex]