zad 1 [latex]a) log_{c}h=g extless = extgreater c^{g}=h\ \ b) log_{c}k+log_{c}p=log_{c}(k*p)[/latex] zad 2 [latex]a) 27^{frac{1}{3}}+5^{-2}=(3^{3})^{frac{1}{3}}+(frac{1}{5})^{2}=3^{3*frac{1}{3}}+frac{1}{25}=3+frac{1}{25}=3frac{1}{25}\ \ b) frac{(17^{4})^{5}:17^{7}}{17^{5}*(17^{3})^{2}}= frac{17^{4*5}:17^{7}}{17^{5}*17^{3*2}}= frac{17^{20}:17^{7}}{17^{5}*17^{6}}=\ \ =frac{17^{20-7}}{17^{5+6}}=17^{13}:17^{11}=17^{13-11}=17^{2}=289[/latex] [latex]c) sqrt[4]{6}*6^{frac{7}{4}}=6^{frac{1}{4}}*6^{frac{7}{4}}=6^{frac{1}{4}+frac{7}{4}}=6^{frac{8}{4}}=6^{2}=36\ \ d) frac{125^{3}:25^{4}}{5^{5}}=frac{(5^{3})^{3}:(5^{2})^{4}}{5^{5}}=frac{5^{9}:5^{8}}{5^{5}}=5^{9-8}:5^{5}=5^{1}:5^{5}=\ \ =5^{1-5}=5^{-4}=frac{1}{625}\[/latex] [latex] e) log_{2}64-log_{7}{7}+log_{3}1=log_{2}2^{6}-1+0=\ \ =6log_{2}2-1=6*1-1=6-1=5\ \ f) log_{6}{180}+log_{6}5=log_{6}(180*5)=log_{6}900=log_{6}(36*25)=\ \ =log_{6}36+log_{6}25=log_{6}6^{2}+log_{6}5^{2}=2log_{6}6+2log_{6}5=\ \ =2+2log_{6}5[/latex] [latex]g) log_{8}8^{11}-12log_{12}15=11log_{8}8-12log_{12}15=11-12log_{12}15\ \ h) log_{5}500-log_{5}4=log_{5}(500:4)=log_{5}125=log_{5}5^{3}=\ \ =3log_{5}5=3[/latex] zad 3 [latex]log_{3}2 approx 0,63\ \ a) log_{3}2^{7}=7log_{3}2 approx 7*0,63 =4,41\ \ b) log_{3}54=log_{3}(27*2)=log_{3}3^{3}+log_{3}2 approx 3+0,63=3,63\ \ c) log_{3}frac{16}{81}=log_{3}16-log_{3}81=log_{3}2^{4}-log_{3}3^{4}=\ \ =4log_{3}2-4log_{3}3 approx 4*0,63-4=2,52-4=-1,48[/latex]
Zad. 1 Uzupełnij zapisy:
a) [latex]log_{c}h=g extless --- extgreater _______________________[/latex]
b) [latex]log_{c}k+log_{c}p= _______________________[/latex]
Zad. 2 Oblicz:
a) [latex]27^{frac{1}{3}}+5^{-2}= [/latex]
b) [latex]frac{(17^{4})^{5}:17^{7}}{17^{5}*(17^{3})^{2}} [/latex]
c) [latex] sqrt[4]{6}*6^{ frac{7}{4}}= [/latex]
d) [latex] frac{125^{3}:25^{4}}{5^{5}}= [/latex]
e) [latex]log_{2}64-log_{7}7+log_{3}1= [/latex]
f) [latex]log_{6}180+log_{6}5= [/latex]
g) [latex]log_{8}8^{11}-12log_{12}15= [/latex]
h) [latex]log_{5}500-log_{5}4= [/latex]
Zad. 3 Oblicz przybliżoną wartość logarytmu,przyjmując , że [latex]log_{3}2 [/latex] w przybliżeniu 0,63
a) [latex]log_{3}2^{7}= [/latex]
b) [latex]log_{3}54= [/latex]
c) [latex]log_{3} frac{16}{81}= [/latex]
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