Question

Вычислите:

а)$2log_{sqrt{5}}25+frac{1}{3}log_{2}64$

б)$4log_{6}216-2log_{0,5}(log_{3}81)$

в)$2log_{2}frac{1}{4}+3log_{frac{1}{3}}27$

г)$5logx_{frac{1}{5}}625+81log_{4}(log_{16}256)$ 

с решением пжл)) 

Answer 1

$2log_{ sqrt{5} } 25 + 1/3log_2 64=2log_{5^{1/2}} } 5^2 + 1/3log_2 2^6=2*2*2+1/3*6=6+2=8;</var><var>$

 

$</var><var>4log_{6}216-2log_{0,5}(log_{3}81)=4log_{6} 6^3 - 2log_{0,5}(log_{3} 3^4)= 4*3 - 2log_{0,5} 4=12 - 2log_{0,5} 0.5^{-2}=12-2*(-2)=12+4=16;</var><var>$

 

$</var><var>2log_{2}frac{1}{4}+3log_{frac{1}{3}}27=2log_{2} 2^{-2}+3log_{frac{1}{3}} frac{1}{3}^{-3}= =2*(-2)+3*(-3)=-4-9=-13;</var><var>$

 

$</var><var>5logx_{frac{1}{5}}625+81log_{4}(log_{16}256) =5logx_{frac{1}{5}} frac{1}{5})^{-4}+81log_{4}(log_{16} 16^2)=5*(-4)+81*log_{4} 2=-20+81*log_{2^2} 2=-20+81/2=-20+40.5=20.5$