1)Oblicz: log 4 9 * log3 1282) Oblicz:a) b) c) d) 3) Wyznacz x jeśli:a) log2 8 + x * log3=log1b)(2x.... Question from @gledzior

November 2018 1 777 Report

1)Oblicz:

log 4 9 * log3 128

2) Oblicz:

a) $3 ^{log3 z liczby 2}$
b) $100 ^{1+log5}$
c) $8 ^{log2 z liczby 5- \frac{1}{3} }$
d) $4 ^{-1+log2 z liczby3}$

3) Wyznacz x jeśli:
a) log2 8 + x * log $_{ \sqrt{3} }$ 3=log1
b)(2x-3)* log4 16=log2 18- log2 9

4) Wiedząc że:

a) log5a i log5 7=b
Oblicz: log125 28

b) log6 2=a i log6 5=b
Oblicz log36 0,8

heh Definicja logarytmu:
$\log_{a}{b}=c\
<=>\ a^{c}=b$

Własności logarytmów:
$\log_{a}{1}=0\\
\log_{a}{a}=1\\ a^{\log_{a}{b}}=b$

Prawa działań na logarytmach:
$log_{a}{b_{1}*b_{2}}=log_{a}{b_{1}}+log_{a}{b_{2}}\\
log_{a}{\frac{b_{1}}{b_{2}}}=log_{a}{b_{1}}-log_{a}{b_{2}}\\
log_{a}{b^{n}}=n*log_{a}{b}\\ log_{a}{b}=\frac{1}{log_{b}{a}}\\
log_{a}{b}=\frac{log_{c}b}{log_{c}a}$
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zad 1
$log_{4}9*log_{3}{128}=log_{4}3^{2}*log_{3}2^{7}=2*log_{4}3*7*log_{3}{2}=\\
=14*\frac{1}{log_{3}4}*log_{3}{2}=14*\frac{1}{log_{3}2^{2}}*log_{3}{2}=\\
=14*\frac{1}{2log_{3}{2}}*log_{3}{2}=14*\frac{1}{2}=7$
============================================
zad 2
$a)\ 3^{log_{3}2}=2$
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$b)\ 100^{1+log5}=100^{1+log_{10}5}=100^{1}*100^{log_{10}5}=100*(10^{2})^{log_{10}5}=\\
=100*10^{2log_{10}5}=100*10^{log_{10}5^{2}}=100*5^{2}=100*25=2500$
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$c)\ 8^{log_{2}5-\frac{1}{3}}=(2^{3})^{log_{2}{5}-\frac{1}{3}}=2^{3*(log_{2}{5}*\frac{1}{3})}=\\
=2^{3log_{2}{5}-1}=2^{log_{2}5^{3}}*2^{-1}=5^{3}*\frac{1}{2}=\\
=125*\frac{1}{2}=\frac{125}{2}=62\frac{1}{2}$
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$d)\ 4^{-1+log_{2}3}=4^{-1}*4^{log_{2}3}=\frac{1}{4}*(2^{2})^{log_{2}3}=\frac{1}{4}*2^{2log_{2}{3}}=\\
=\frac{1}{4}*2^{log_{2}3^{2}}=\frac{1}{4}*3^{2}=\frac{1}{4}*9=\frac{9}{4}=2\frac{1}{4}$
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zad 3
$a)\ log_{2}8+x*log_{\sqrt{3}}3=log1\\
log_{2}2^{3}+x*\frac{1}{log_{3}\sqrt{3}}=0\\
3*log_{2}{2}+x*\frac{1}{log_{3}3^{\frac{1}{2}}}=0\\
3*1+x*\frac{1}{\frac{1}{2}*log_{3}{3}}=0\\
3+x*\frac{1}{\frac{1}{2}*1}=0\\
3+x*1*\frac{2}{1}=0\\
3+2x=0\\
2x=-3\\
x=\frac{-3}{2}\\
x=-1\frac{1}{2}$
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$b)\ (2x-3)log_{4}{16}=log_{2}{18}-log_{2}{9}\\
(2x-3)log_{4}4^{2}=log_{2}(18:9)\\
(2x-3)*2*log_{4}{4}=log_{2}2\\
(2x-3)*2*1=1\\
4x-6=1\\
4x=7\\
x=\frac{7}{4}\\
x=1\frac{3}{4}$
============================================
zad 4
$a)\ log_{125}{28}=\frac{1}{log_{28}125}=\frac{1}{log_{28}5^{3}}=\frac{1}{3*log_{28}{5}}=\\ =\frac{1}{3}*\frac{1}{log_{28}5}=\frac{1}{3}*log_{5}28=\frac{1}{3}*log_{5}(4*7)=\\ =\frac{1}{3}*(log_{5}4+log_{5}7)$
[W treści zapisu przykładu jest błąd]
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$b)\ log_{36}0,8=\frac{1}{log_{0,8}36}=\frac{1}{log_{0,8}6^{2}}=\frac{1}{2*log_{0,8}6}=\\
=\frac{1}{2}*\frac{1}{log_{0,8}6}=\frac{1}{2}*log_{6}0,8=\frac{1}{2}*log_{6}\frac{8}{10}=\\
=\frac{1}{2}*log_{6}\frac{4}{5}=\frac{1}{2}*(log_{6}4-log_{6}5)=\frac{1}{2}*(log_{6}2^{2}-log_{6}5)=\\
=\frac{1}{2}*(2*log_{6}2-log_{6}5)=\frac{1}{2}*(2a-b)=a-\frac{1}{2}b$

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