Matematyka- zadania w załączniku.... Question from @derrapes

derrapes @derrapes

October 2018 1 143 Report

Matematyka- zadania w załączniku.

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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$\log_{\frac{1}{a}}{b}=\log_{a}{\frac{1}{b}}\\ L=\log_{\frac{1}{a}}{b}=\frac{1}{\log_{b}\frac{1}{a}}=\frac{1}{log_{b}{a^{-1}}}=\frac{1}{-\log_{b}{a}}=\\ =-\frac{1}{\frac{1}{log_{a}{b}}}=-1*\log_{a}{b}=\log_{a}{b^{-1}}=\log_{a}{\frac{1}{b}}=P$

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$\log_{\sqrt{a}}{b}=\log_{a}{b^{2}}\\ L=\log_{\sqrt{a}}{b}=\frac{1}{\log_{b}{\sqrt{a}}}=\frac{1}{\log_{b}a^{\frac{1}{2}}}=\frac{1}{\frac{1}{2}\log_{b}{a}}=\\ =\frac{1}{\frac{1}{2log_{a}{b}}}=1*2\log_{a}{b}=\log_{a}{b^{2}}=P$

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$\log_{a^{2}}{b}=\log_{a}{b^{\frac{1}{2}}}\\ L=\log_{a^{2}}{b}=\frac{1}{\log_{b}{a^{2}}}=\frac{1}{2\log_{b}{a}}=\frac{1}{2}*\log_{a}{b}=\log_{a}{b^{\frac{1}{2}}}=P$

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$\log_{b}{\sqrt[n]{a}}=n\log_{a}{b}\\ L=\log_{b}{\sqrt[n]{a}}=\log_{b}{a^{\frac{1}{n}}}=\frac{1}{n}\log_{b}{a}=\frac{1}{nlog_{a}{b}}=\\ =\frac{1}{log_{a}{b^{n}}}=\log_{b^{n}}{a} \neq P$