Logarytmy

$$\begin{gathered} a^{{\log }_{a}b}=b, \\ \text{gdzie}a\in {\mathbb{R}}^{+}\setminus \left\{1\right\},b\in {\mathbb{R}}^{+} \end{gathered}$$

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a)

$5^{3-{\log }_{5}2}=5^3\cdot 5^{-{\log }_{5}2}=5^3\cdot 5^{\left(-1\right)\cdot {\log }_{5}2}=5^3\cdot {\left(5^{{\log }_{5}2}\right)}^{-1}=125\cdot {\left(2\right)}^{-1}=125\cdot \frac{1}{2}=62\frac{1}{2}$

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b)

$5^{2-{\log }_{5}9}-3^{-1+{\log }_{3}18}=5^2\cdot 5^{-{\log }_{5}9}-3^{-1}\cdot 3^{{\log }_{3}18}=25\cdot {\left(5^{{\log }_{5}9}\right)}^{-1}-\frac{1}{3}\cdot 18=25\cdot 9^{-1}-6=\frac{25}{9}-6=-3\frac{2}{9}$

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c)

${\left(\frac{1}{3}\right)}^{{\log }_{3}7}={\left(3^{-1}\right)}^{{\log }_{3}7}={\left(3^{{\log }_{3}7}\right)}^{-1}=7^{-1}=\frac{1}{7}$

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d)

$49^{{\log }_{7}5}+27^{{\log }_{3}4}+2^{{\log }_{2}5}={\left(7^2\right)}^{{\log }_{7}5}+{\left(3^3\right)}^{{\log }_{3}4}+5={\left(7^{{\log }_{7}5}\right)}^2+{\left(3^{{\log }_{3}4}\right)}^3+5=5^2+4^3+5=25+64+5=30+64=94$

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e)

${\left(\sqrt{5}\right)}^{{\log }_{5}16}-{\left(\sqrt{7}\right)}^{{\log }_{7}6}={\left(5^{\frac{1}{2}}\right)}^{{\log }_{5}16}-{\left(7^{\frac{1}{2}}\right)}^{{\log }_{5}6}={\left(5^{{\log }_{5}16}\right)}^{\frac{1}{2}}-{\left(7^{{\log }_{7}6}\right)}^{\frac{1}{2}}=16^{\frac{1}{2}}-6^{\frac{1}{2}}=4-\sqrt{6}$

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f)

$25^{3-{\log }_5\sqrt{15}}=25^3\cdot {\left(5^2\right)}^{-{\log }_5\sqrt{15}}={\left(5^2\right)}^3\cdot {\left(5^{{\log }_5\sqrt{15}}\right)}^{-2}=5^6\cdot {\left(\sqrt{15}\right)}^{-2}=5^6\cdot \frac{1}{15}=5^5\cdot \frac{1}{3}=\frac{3125}{3}=1041\frac{2}{3}$