Question

Dacă se poate si vreti as dori si explicatie

Answer 1

a)
[tex]log_3(log_2(log_4256))=log_3(log_2(log_44^{4}))=\\\\ =log_3(log_24)=log_3(log_22^{2})=log_32[/tex]

b)
[tex]log_{ \frac{1}{2} }4+log_81024=log_{2^{-1}}4+log_{2^{3}}2^{10}=\\\\ =-1log_{2}2^{2}+3log_{2}2^{10}=-1*2+3*10=30-2=28[/tex]

c)
$4^{3-log_42}=4^{3-2}=4^1=4$

Answer 2

b) 

[tex]\it log_{\frac{1}{2}}4= \dfrac{log_2 4}{log_2{\dfrac{1}{2}}} =\dfrac{log_22^2}{log_22^{-1}} =\dfrac{2}{-1} = -2 [/tex]

$\it log_8 1024 = \dfrac{log_21024}{log_28} =\dfrac{log_2 2^{10}}{log_22^3} = \dfrac{10}{3}$

$\it-2+\dfrac{10}{3} = \dfrac{-6+10}{3} =\dfrac{4}{3}\ .$

c)

$\it \log_42 = log_4\sqrt4 = log_4 4^{\dfrac{1}{2}} = \dfrac{1}{2}\cdot log_4 4 = \dfrac{1}{2}$