Question

log in baza 2 din 24/log in baza 96 din 2 - log in baza 2 din 192/log in baza 12 din 2​

Answer 1

 

$\displaystyle\\\text{Folosim formulele:}\\\\\log_ab=\frac{1}{\log_ba}\\\\\log_a{(b\times c)}=\log_ab+\log_ac\\\\\text{Rezolvare:}\\\\\frac{\log_2{24}}{\log_{96}2}-\frac{\log_2{192}}{\log_{12}2}=\\\\\\=\frac{~~\log_2{24}~~}{\dfrac{1}{\log_2{96}}}-\frac{~~\log_2{192}~~}{\dfrac{1}{\log_2{12}}}=\\\\\\=\log_2{(24)}\times\log_2{(96)}-\log_2{(192)}\times\log_2{12}=$

$\displaystyle\\=\log_2{(3\times8)}\times\log_2{(3\times32)}-\log_2{(3\times64)}\times\log_2{3\times4}=\\\\=\Big(\log_23+\log_28\Big)\Big(\log_23 + \log_2{32}\Big)-\Big(\log_23+\log_2{64}\Big)\Big(\log_23+\log_24\Big)=\\\\=\Big(\log_23+\log_22^3\Big)\Big(\log_23 + \log_22^5\Big)-\Big(\log_23+\log_22^6\Big)\Big(\log_23+\log_22^2\Big)=\\\\=\Big(\log_23+3\Big)\Big(\log_23 + 5\Big)-\Big(\log_23+6\Big)\Big(\log_23+2\Big)=$

$\displaystyle\\\\=\Big(\log_23\Big)^2+3\log_23+5\log_23 + 3\times5-\left[\Big(\log_23\Big)^2+6\log_23+2\log_23+6\times2\right]=\\\\=\Big(\log_23\Big)^2+8\log_23+15-\left[\Big(\log_23\Big)^2+8\log_23+12\right]=\\\\=\underline{\underline{\Big(\log_23\Big)^2}}+\underline{8\log_23}+15-\underline{\underline{\Big(\log_23\Big)^2}}-\underline{8\log_23}-12=\\\\=15-12=\boxed{\Big~~\bf3~~}$

.