Zadanie 10, bardzo prosze o pomoc ; ).... Question from @Kenth

November 2018 1 26 Report

Zadanie 10, bardzo prosze o pomoc ; )

Roma A)
$log_5 \ 5^{-\frac{2}{7}} = -\frac{2}{7} \cdot log_5 \ 5= -\frac{2}{7} \cdot 1 =-\frac{2}{7}$
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$log_5 \ (5^3)^7 = log_5 \ 5^{3 \cdot 7}= log_5 \ 5^{21} =21 \cdot log_5 \ 5 =21 \cdot 1 = 21$
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$log_5 \ (5^{\frac{2}{3}})^{-6} = log_5 \ 5^{\frac{2}{3} \cdot (-6)}=log_5 \ 5^{-4} = - 4 \cdot log_5 \ 5 = -4 \cdot 1 = -4$
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$log_5 \ (5^{-4})^{\frac{1}{5}}= log_5 \ 5^{- 4 \cdot \frac{1}{5}}=log_5 \ 5^{-\frac{4}{5}} = - \frac{4}{5} \cdot log_5 \ 5 = -\frac{4}{5} \cdot 1 = -\frac{4}{5}$

b)
$log_6 \ (6^{15})^{-4}=log_6 \ 6^{15 \cdot (-4)}=log_6 \ 6^{-60}=-60 \cdot log_6 \ 6 = - 60\cdot 1 = \\\ = - 60$
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$log_6 \ \frac{1}{36} =log_6 \ \frac{1}{6^2}=log_6 \ 6^{-2}=-2 \cdot log_6 \ 6 = - 2\cdot 1 = -2$
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$log_6 \ (\sqrt[3]{6})^2 =log_6 \ (6^{\frac{1}{3}})^2=log_6 \ 6^{\frac{1}{3} \cdot 2}= log_6 \ 6^{\frac{2}{3}}=\frac{2}{3} \cdot log_6 \ 6 = \\\
= \frac{2}{3} \cdot 1 =\frac{2}{3}$
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$log_6 \ \frac{1}{\sqrt[5]{36}}=log_6 \ \frac{1}{(6^2)^{\frac{1}{5}}}= log_6 \ \frac{1}{6^{2 \cdot\frac{1}{5}}}= log_6 \ \frac{1}{6^{\frac{2}{5}}}=log_6 \ 6^{-\frac{2}{3}}=\\\
=-\frac{2}{5} \cdot log_6 \ 6 = -\frac{2}{5} \cdot 1 =-\frac{2}{5}$

c)
$log_{\frac{1}{2}} \ 2^{-4}=log_{\frac{1}{2}} \ (\frac{1}{2})^4= 4\cdot log_{\frac{1}{2}} \ \frac{1}{2} =4 \cdot 1 = 4$
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$log_{\frac{1}{2}} \ ((0,5)^7)^8=log_{\frac{1}{2}} \ (0,5)^{7\cdot 8}= log_{\frac{1}{2}} \ (0,5)^{56}=56 \cdot log_{\frac{1}{2}} \ 0,5 =\\\
=56 \cdot 1 =56$
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$log_{\frac{1}{2}} \ 4^{13}=log_{\frac{1}{2}} \ (2^2)^{13}= log_{\frac{1}{2}} \ 2^{2 \cdot 13}= log_{\frac{1}{2}} \ 2^{26}=log_{\frac{1}{2}} \ (\frac{1}{2})^{-26}=\\\
=-26 \cdot log_{\frac{1}{2}} \ \frac{1}{2} =-26 \cdot 1 =-26$
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$log_{\frac{1}{2}} \ (\sqrt{2})^{18}=log_{\frac{1}{2}} \ (2^{\frac{1}{2}})^{18}= log_{\frac{1}{2}} \ 2^{\frac{1}{2} \cdot 18}= log_{\frac{1}{2}} \ 2^9=log_{\frac{1}{2}} \ (\frac{1}{2})^{-9}=\\\
=-9 \cdot log_{\frac{1}{2}} \ \frac{1}{2} =-9 \cdot 1 =-9$

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Skorzystano z własności logarytmów:
$log_a \ b^r = r \cdot log_a \ b \\\\
log_a \ a= 1$
oraz własności potęg i pierwiastków:
$(a^m)^n = a^{m \cdot n} \\\\
a^{-n} = \frac{1}{a^n} \\\\
\sqrt[n]{a} = a^{\frac{1}{n}}$

1 votes Thanks 1

Kenth Dziękuje bardzo :)