dla log2 = 0,3 i log5 = 0,7:
a)
[tex]\log20=\log(4\cdot5)=\log(2^2\cdot5)=\log2^2+\log5=2\log2+\log5=\\\\{}\qquad\qquad\qquad\qquad\qquad=2\cdot0,3+0,7=1,3\\\\\log50=\log(25\cdot2)=\log(5^2\cdot2)=\log5^2+\log2=2\log5+\log2=\\\\{}\qquad\qquad\qquad\qquad\qquad=2\cdot0,7+0,3=1,7\\\\ \log0,2=\log\frac1{5}=\log5^{-1}=-1\cdot0,7 =-0,7\\\\ \log0,4=\log\frac4{10}=\log2^2-\log10=2\log2-\log10=2\cdot0,3-1=-0,4\\\\ \log0,04=\log\frac4{100}=\log\frac{2^2}{10^2}=\log2^2-\log10^2=2\log2-2\log10=\\\\{}\qquad\qquad\qquad\qquad\qquad=2\cdot0,3-2\cdot1=-1,4[/tex]
[tex]\log\sqrt{20}=\log(\sqrt4\cdot\sqrt5)= \log(2\cdot5^\frac12)= \log2+\log\big5^\frac12 =\log2+\frac12\log5=\\\\ {}\qquad\qquad\qquad\qquad\qquad\qquad=0,3+\frac12\cdot0,7=0,3+0,35=0,65[/tex]
b)
[tex]\log\frac25=\log2-\log5=0,3-0,7=-0,4\\\\\log\frac45=\log\frac{2^2}5=\log2^2-\log5=2\log2-\log5=2\cdot0,3-0,7=-0,1\\\\\log\frac{25}{64}=\log\frac{5^2}{2^6}=\log5^2-\log2^6=2\log5-6\log2=2\cdot0,7-6\cdot0,3=-0,4\\\\ \log2,5=\log\frac52=\log5-\log2=0,7-0,3=0,4\\\\\log6,25=\log\frac{625}{100}=\log\frac{5^4}{10^2}=\log5^4-\log10^2=4\log5-2\log10=\\\\{}\qquad\qquad\qquad\qquad\qquad\qquad\qquad=4\cdot0,7-2\cdot1=0,8[/tex]
[tex]\log25\sqrt2=\log(5^2\cdot2^\frac12)=\log5^2+\log2^\frac12=2\log5+\frac12\log2=\\\\{}\qquad\qquad\qquad\qquad\qquad=2\cdot0,7+\frac12\cdot0,3=1,4+0,15=1,55[/tex]
