Proszę o rozwiązanie. Odpowiedz ma być czytelna.... Question from @Mariuszkam

December 2018 1 6 Report

Proszę o rozwiązanie. Odpowiedz ma być czytelna

unicorn05 1.

$log_264=c\ \ \Leftrightarrow \ 2^c=64\ \Leftrightarrow \ 2^c=2^6\ \Leftrightarrow \ c=6\ \ \Rightarrow \ log_264=6$

$log_2512=c\ \ \Leftrightarrow \ 2^c=512\ \Leftrightarrow \ 2^c=2^9\ \Leftrightarrow \ c=9\ \ \Rightarrow \ log_2512=9$

$log_{0,5}128=c\ \ \Leftrightarrow \ (\frac{1}{2})^c=128\ \Leftrightarrow \ 2^{-c}=2^7\ \Leftrightarrow \ -c=7\ \Leftrightarrow \ c=-7\ \\ \ log_{0,5}128=-7$

$log_{\frac{1}{2}}16=c\ \ \Leftrightarrow \ (\frac{1}{2})^c = 16\ \Leftrightarrow \ (\frac{1}{2})^c = 2^4\ \ \Leftrightarrow \ (\frac{1}{2})^c = (\frac{1}{2})^{-4} \ \Leftrightarrow \ c= -4\\ log_{\frac{1}{2}}16=-4$

$log_{0,1}1000=c\ \ \ \Leftrightarrow \ \ (0,1)^c = 1000 \ \Leftrightarrow \ (\frac{1}{10})^c = (\frac{1}{10})^{-3} \Leftrightarrow \ c=-3\\ log_{0,1}1000=-3
$

$log100000=c\ \ \ \Leftrightarrow \ \ 10^c = 100000\ \Leftrightarrow \ 10^c=10^5\ \Leftrightarrow \ c=5 \\ log100000=5$

$log_{0,4}{2,5}=c\ \ \ \Leftrightarrow \ \ (\frac{4}{10})^c=2,5\ \Leftrightarrow (\frac{10}{4})^{-c}=2,5\ \Leftrightarrow \ -c=1\Leftrightarrow c=-1 \\ log_{0,4}{2,5}=-1$

$log_50,2=c\ \ \ \Leftrightarrow \ \ 5^c = \frac{2}{10}\ \Leftrightarrow \ 5^c = \frac{1}{5}\ \Leftrightarrow \ 5^c = 5^{-1}\ \Leftrightarrow \ c = -1 \\ log_50,2=-1$

$log_ {\sqrt{2}}16 =c\ \ \ \Leftrightarrow \ \ (\sqrt{2})^c=16\ \Leftrightarrow \ 2^{ \frac{1}{2} c}=2^4\ \Leftrightarrow \ \frac{1}{2} c=4\ \Leftrightarrow \ \ c=8 \\ log_ {\sqrt{2}}16 =8$

$log_5 \sqrt{125} = c \ \ \ \Leftrightarrow \ \ 5^c = \sqrt{125} \ \Leftrightarrow \ 5^c=(5^3)^{\frac{1}{2}} \Leftrightarrow \ 5^c=5^{\frac{3}{2}} \Leftrightarrow \ c={\frac{3}{2}} \\ log_5 \sqrt{125} =\frac{3}{2}}$

$log_71 = c \ \ \ \Leftrightarrow \ \ 7^c=1 \ \ \ \Leftrightarrow \ \ c =0 \ \ \Rightarrow \ log_71 =0$

$log_6{ \sqrt{6}} = log_6(6^{\frac{1}{2}})= \frac{1}{2}log_66 =\frac{1}{2}*1=\frac{1}{2}$

2.
$2^{log_29}=9$

$\frac{1}{2} ^{log_211} = (2^{-1})^{log_211} = 2^{-log_211}= 2^{log_211^{-1}}=11^{-1}= \frac{1}{11}$

$4^{log_27} = (2^2)^{log_27}=2^{2log_27}=2^{log_27^2}=7^2=49$

$(\sqrt{2}\ ) ^{log_26}=(2^{ \frac{1}{2} })^{log_26}=2^{ \frac{1}{2} log_26}=2^{log_26^{\frac{1}{2}}}=6^{\frac{1}{2}}= \sqrt{6}$

3
a)
$log125 + log4-log5 = log \frac{125*4}{5}= log {25*4}= log100 = 2$
(bo 10² =100)

b)
$log_336 -log_32+log_3{ \frac{1}{6} }=log_3( \frac{36}{2} * \frac{1}{6} )=log_3 \frac{6}{2}=log_33=1$

c)
$log2+log50=log(2*50)=log100=2$ (bo 10² =100)

d)
$log_318-log_32 = log_3{ \frac{18}{2} }=log_39=log_33^2=2log_33=2*1=2$

e)
$log_{14}49+2log_{14}2=log_{14}7^2+2log_{14}2=2log_{14}7+2log_{14}2= \\ \\2(log_{14}7+log_{14}2) = 2log_{14}(7*2)= 2log_{14}14=2*1=2$

f)
$27^{log_32}=(3^3)^{log_32}=3^{3log_32}=3^{log_32^3}=2^3 =8$

g)
$2^{log_25}=5$

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