Вычислите: а) $$log_{2}16 + log_{\frac{1}{3}}9$$; б) $$5^{log_{5}10-1}$$; в) $$log_{6}9 + 2log_{6}2$$; г) $$lg \sqrt{30} - lg \sqrt{3}$$

a) $$log_{2}16 + log_{\frac{1}{3}}9$$

$$log_{2}16 = 4$$, так как $$2^4 = 16$$

$$log_{\frac{1}{3}}9 = -2$$, так как $$(\frac{1}{3})^{-2} = 3^2 = 9$$

Следовательно, $$log_{2}16 + log_{\frac{1}{3}}9 = 4 + (-2) = 2$$

б) $$5^{log_{5}10-1}$$

$$5^{log_{5}10-1} = 5^{log_{5}10} \cdot 5^{-1} = \frac{5^{log_{5}10}}{5} = \frac{10}{5} = 2$$

в) $$log_{6}9 + 2log_{6}2$$

$$log_{6}9 + 2log_{6}2 = log_{6}9 + log_{6}2^2 = log_{6}9 + log_{6}4 = log_{6}(9 \cdot 4) = log_{6}36 = 2$$

г) $$lg \sqrt{30} - lg \sqrt{3}$$

$$lg \sqrt{30} - lg \sqrt{3} = lg \frac{\sqrt{30}}{\sqrt{3}} = lg \sqrt{\frac{30}{3}} = lg \sqrt{10} = lg 10^{\frac{1}{2}} = \frac{1}{2} lg 10 = \frac{1}{2} \cdot 1 = \frac{1}{2}$$

Ответ: a) 2; б) 2; в) 2; г) 1/2