Question

Dau coroană , urgent

Answer 1

$\displaystyle \mathtt{c)~log_{12}3+log_{12}4+12^{log_{144}4}+log_{ \frac{1}{2} }8}=\\ \\ \mathtt{=log_{12}(3 \cdot 4)+12^{log_{12^2}4}+log_{2^{-1}}8=log_{12}12+12^{^{ \frac{1}{2}log_{12} 4}}-log_28=}\\ \\ \mathtt{=1+12^{log_{12}4^ \frac{1}{2} }-log_22^3=1+4^{ \frac{1}{2} }-3log_22=1+ \sqrt{4}-3=1+2-3=0 }$

$\displaystyle\mathtt{f)~2lg5+ \frac{1}{2}lg16=lg5^2+lg16^{ \frac{1}{2} }=lg25+lg \sqrt{16}=lg25+lg4=}\\ \\ \mathtt{=lg(25 \cdot 4)=lg100=lg10^2=2lg10=2}$

$\displaystyle \mathtt{i)~log_{ \sqrt{3} }18-log_34=log_{3^{ \frac{1}{2} }}18-log_34= \frac{1}{ \frac{1}{2} } log_318-log_34=}\\ \\ \mathtt{=2log_318-log_34=log_318^2-log_34=log_3324-log_34=}\\ \\ \mathtt{=log_3 \frac{324}{4} =log_381=log_33^4=4log_33=4}$

$\displaystyle \mathtt{l)~3log_36- \frac{3}{2} log_34=log_36^3- \frac{3}{2} log_32^2=log_3216-\not2 \cdot \frac{3}{\not2} log_32=}\\ \\ \mathtt{=log_3216-3log_32=log_3216-log_32^3=log_3216-log_38=}\\ \\ \mathtt{=log_3 \frac{216}{8}=log_327=log_33^3=3log_33=3 }$