Question

vă rog îmi trebuie repede va rog dau coroană ​

Answer 1

 

$\displaystyle\bf\\3)\\\\a)~log_2~8=log_2~2^3=3~~~sau~~~log_2~2^3=3log_2~2=3\times1=3\\\\b)~log_4~1024=log_4~4^5=5\\\\c)~log_3~27=log_3~3^3=3\\\\d)~\Big(\sqrt{3}\Big)^2=3~\implies~log_{\sqrt{3}}~3=log_{\sqrt{3}}~\Big(\sqrt{3}\Big)^2=2\\\\e)~log_2~\sqrt{8}=log_2~8^\frac{1}{2}=log_2~\Big(2^3\Big)^\frac{1}{2}=log_2~\Big(2\Big)^{3\times\frac{1}{2}}=log_2~\Big(2\Big)^{\frac{3}{2}}=\frac{3}{2}$

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$\displaystyle\bf\\f)~log_{0,5}~4=log_{\frac{1}{2}}~2^2=log_{\frac{1}{2}}~\frac{1}{2^{-2}}=log_{\frac{1}{2}} ~\left(\frac{1}{2}\right)^{-2}=-2\\\\g)~log_3~\frac{1}{81}=log_3~\frac{1}{3^4}=log_3~3^{-4}=-4\\\\h)~lg~100=log_{10}~100=log_{10}~10^2=2\\\\i)~ln~e^3=log_e~e^3=3\\\\j)~log_{\sqrt[3]{4}}~8=log_{2^\frac{2}{3} }~2^3=\frac{3}{2}\times3\times log_2 2 =\frac{3}{2}\times3\times1=\frac{9}{2}$.

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$\displaystyle\bf\\k)~log_9~0,(3)=log_{3^2}~\frac{1}{3}=log_{3^2}~3^{-1}=\\\\=\frac{1}{2}\times(-1)\times log_3~3=\frac{1}{2}\times(-1)\times1=-\frac{1}{2}\\\\l)~lg~0,01=log_{10}~\frac{1}{100}=log_{10}~\frac{1}{10^2}=log_{10}~10^{-2}=-2$