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[Гост]

log4 (49 ).log7 (25).log5(16)+ log8 (13)/log2(1/13)

[ammornil]

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[tex]log_{a}P^{n}=n.log_{a}P, \>\phantom{\ \ \ } log_{a}P=\frac{log_{b}P}{log_{b}a}, \>\phantom{\ \ \ } log_{a}a^{n}=n, \>\phantom{\ \ \ } log_{a}a=1, \>\phantom{\ \ \ } log_{a}{b}=\frac{1}{log_{b}{a}}[/tex]

[tex]log_{a^{n}}P=\frac{log_{a}P}{log_{a}a^{n}}=\frac{log_{a}P}{n},\>\phantom{\ \ \ } log_{a^{n}}{b^{n}}=log_{a}{b}[/tex]

[tex]A=log_{4}49.log_{7}25.log_{5}16+\frac{log_{8}13}{log_{2}\frac{1}{13}}=?[/tex]

[tex]log_{4}{49}=log_{2^{2}}{7^{2}}=\frac{1}{2}.log_{2}{7^{2}}=\frac{1}{2}.2.log_{2}{7}=log_{2}{7}[/tex]

[tex]log_{7}{25}=log_{7}{5^{2}}=2log_{7}{5}=2\frac{log_{2}{5}}{log_{2}{7}}[/tex]

[tex]log_{5}{16}=log_{5}{2^{4}}=4log_{5}{2}=4\frac{log_{2}{2}}{log_{2}{5}}=\frac{4}{log_{2}{5}}[/tex]

[tex]log_{8}{13}=log_{2^{3}}{13}=\frac{1}{3}log_{2}{13}[/tex]

[tex]log_{2}{\frac{1}{13}}=log_{2}{13^{-1}}=-log_{2}{13}[/tex]

[tex]A=\cancel{log_{2}{7}}.2\frac{\cancel{log_{2}{5}}}{\cancel{log_{2}{7}}}.\frac{4}{\cancel{log_{2}{5}}}+\frac{\frac{1}{3}\cancel{log_{2}{13}}}{-\cancel{log_{2}{13}}}=8-\frac{1}{3}=7\frac{2}{3}[/tex]

[Гост]

Благодаря!

log4 (49 ).log7 (25).log5(16)+ log8 (13)/log2(1/13)