Ответ:
Применяем свойства логарифмов: [tex]a > 0\, ,a\ne 1\ ,\ b > 0\ ,\ c > 0[/tex] ,
[tex]log_{a}a=1\ ,\ \ k\cdot log_{a}b=log_{a}b^{k}\ ,\ \ log_{a}(bc)=log_{a}b+log_{a}c[/tex] .
[tex]1)\ \ \dfrac{1}{4}=\dfrac{1}{4}\cdot \underbrace{log_{10}\, 10}_{1}=log_{10}\, 10^{\frac{1}{4}}=log_{10}\, \sqrt[4]{10}=lg\sqrt[4]{10}\\\\\\2)\ \ log_{x}}\, y=5\ \ \ ,\ \ \ \ x > 0\ ,\ x\ne 1\ ,\ y > 0\\\\log_{x}\sqrt[3]{xy}=log_{x}(\sqrt[3]{x}\cdot \sqrt[3]{y})=log_{x}\, x^{\frac{1}{3}}+log_{x}\, y^{\frac{1}{3}}=\dfrac{1}{3}\cdot \underbrace{log_{x}\, x}_{1}+\dfrac{1}{3}\cdot log_{x}\, y=\\\\=\dfrac{1}{3}+\dfrac{1}{3}\cdot 5=\dfrac{6}{3}=2[/tex]
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Ответ:
Применяем свойства логарифмов: [tex]a > 0\, ,a\ne 1\ ,\ b > 0\ ,\ c > 0[/tex] ,
[tex]log_{a}a=1\ ,\ \ k\cdot log_{a}b=log_{a}b^{k}\ ,\ \ log_{a}(bc)=log_{a}b+log_{a}c[/tex] .
[tex]1)\ \ \dfrac{1}{4}=\dfrac{1}{4}\cdot \underbrace{log_{10}\, 10}_{1}=log_{10}\, 10^{\frac{1}{4}}=log_{10}\, \sqrt[4]{10}=lg\sqrt[4]{10}\\\\\\2)\ \ log_{x}}\, y=5\ \ \ ,\ \ \ \ x > 0\ ,\ x\ne 1\ ,\ y > 0\\\\log_{x}\sqrt[3]{xy}=log_{x}(\sqrt[3]{x}\cdot \sqrt[3]{y})=log_{x}\, x^{\frac{1}{3}}+log_{x}\, y^{\frac{1}{3}}=\dfrac{1}{3}\cdot \underbrace{log_{x}\, x}_{1}+\dfrac{1}{3}\cdot log_{x}\, y=\\\\=\dfrac{1}{3}+\dfrac{1}{3}\cdot 5=\dfrac{6}{3}=2[/tex]