Question

Logaritmi : Nu prea știu cum sa fac ex 1. Ma puteți ajuta? (dacă mi-ați putea și explica)

Answer 1

 

$\displaystyle\bf\\ 1)\\\\a)\\\\log_{12}(288)-log_{12}(2)+log_2(lg(100))=\\\\=log_{12}\left(\frac{288}{2}\right)+log_2(2)\\\\\\=log_{12}(144)+log_2(2)=\\\\=log_{12}(12^2)+log_2(2)=2+1=\boxed{\bf3}$

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$\displaystyle\bf\\b)\\\\log_3\left(9\sqrt[\b4]{27\sqrt[\b5]{81} } \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^3\sqrt[\b5]{3^4} } \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^3\times3^\dfrac{4}{5}} \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^3\times3^\dfrac{4}{5}} \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^{\b3+\dfrac{4}{5}}} \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^{\dfrac{3\times5+4}{5}}} \right)=\\\\\\=log_3\left(3^2\sqrt[\b4]{3^{\dfrac{19}{5}}} \right)=$

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$\displaystyle\bf\\=log_3\left(3^\b2}\times3^{\dfrac{19}{5\times4}}\right)=\\\\\\=log_3\left(3^\b2}\times3^{\dfrac{19}{20}}\right)=\\\\\\=log_3\left(3^{\dfrac{19}{20}+2}\right)=\\\\\\=log_3\left(3^{\dfrac{19+2\times20}{20}}\right)=\\\\\\=log_3\left(3^{\dfrac{59}{20}}\right)=\\\\\\\frac{59}{20}log_3\Big(3\Big)=\frac{59}{20}\times1=\boxed{\bf\frac{59}{20}}$