Question

Номер 3,10,6. Алгебра 10-11 класс​

Answer 1

$3^{1+log_{3}5}=3*3^{log_{3}5}=3*5=15\\\\5^{2-log_{5}3}=5^{2}*5^{log_{5}\frac{1}{3}}=25*\frac{1}{3}=8\frac{1}{3}\\\\7^{2log_{7}5}=(7^{log_{7}5})^{2}=5^{2}=25$

$log_{3}27\sqrt[5]{9}=log_{3}(3^{3} *3^{\frac{2}{5}})=log_{3}3^{3,4}=3,4\\\\32^{log_{2}3}=(2^{5})^{log_{2}3}=(2^{log_{2}3})^{5}=3^{5}=243\\\\log_{\frac{1}{2}}\frac{\sqrt{2}}{64}}=log_{\frac{1}{2}} \frac{2^{\frac{1}{2}}}{2^{6}}=log_{\frac{1}{2}}2^{-5,5}=log_{2}2^{5,5} =5,5$

$log_{5}\frac{125x^{4}}{\sqrt[3]{x^{4}}}=log_{5}\frac{5^{3}*x^{4}}{x^{\frac{4}{3}}}=log_{5}5^{3} +log_{5}x^{4}-log_{5}x^{\frac{4}{3}}=3+4log_{5}x-\frac{4}{3}log_{5}x=3+2\frac{2}{3}log_{5}x$