Prosze o pomoc
logarytmy


Prosze O Pomoc Logarytmy class=

Odpowiedź :

[tex]a)\\\log_3306+\log_312-\log_3136=\log_3\dfrac{306\cdot12}{136} = \log_3\dfrac{306\cdot3}{34}=\log_3(9\cdot3)=\\\\ =\log_33^3=3\log_33 = 3\cdot1=3 \\\\\\b)\\\log_28^\frac32+\log_\frac139^{-2}-\log_\frac12(\log100^8) = \\ \\ =\log_2(2^3)^\frac32+\log_\frac13(\frac19)^2-\log_\frac12[\log(10^2)^8]= \\\\ =\log_22^\frac92+\log_\frac13[(\frac13)^2]^2-\log_\frac12(\log10^{16})= \\\\ =\frac92\log_22+\log_\frac13(\frac13)^4-\log_\frac12(16\log10)= \\\\ =\frac92\cdot1+4\log_\frac13\frac13-\log_\frac1216=[/tex]

[tex]=\frac92+4\cdot1-\log_\frac122^4= \\\\ =4\frac12+4-\log_\frac12(\frac12)^{-4}= \\\\ =8\frac12-(-4)\log_\frac12\frac12=\\\\=8\frac12+4=\\\\=12\frac12[/tex]

[tex]c)\\\\4\log_\frac1214+8\log_\frac126-4\log_\frac1263= 4(\log_\frac1214+2\log_\frac126-\log_\frac1263)= \\\\ = 4(\log_\frac1214+\log_\frac126^2-\log_\frac1263)= 4\log_\frac12\dfrac{14\cdot6^2}{63} = 4\log_\frac12\dfrac{2\cdot36}{9}=\\\\= 4\log_\frac12(2\cdot4)= 4\log_\frac122^3= 4\log_\frac12(\frac12)^{-3}= 4\cdot(-3)\cdot\log_\frac12\frac12=-12[/tex]