Упр.43.19 ГДЗ Мордкович 10-11 класс (Алгебра)

Используем свойство логарифма: $$\log_a x+\log_a y=\log_a(xy).$$

1. $$\log_{\sqrt2}\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt2}\left(2\cos\frac{\pi}{8}\right)=\log_{\sqrt2}\left(2\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right).$$

   Так как $$2\sin\frac{\pi}{8}\cos\frac{\pi}{8}=\sin\frac{\pi}{4},$$ то

   $$\log_{\sqrt2}\left(\sin\frac{\pi}{4}\right)=\log_{\sqrt2}\left(\frac{1}{\sqrt2}\right)=\log_{\sqrt2}\left((\sqrt2)^{-1}\right)=-1.$$

2. $$\log_{\frac12}\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)+\log_{\frac12}\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)$$

   $$=\log_{\frac12}\left(\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)\right)$$

   $$=\log_{\frac12}\left(\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{6}\right)=\log_{\frac12}\left(\cos\frac{\pi}{3}\right)=\log_{\frac12}\left(\frac12\right)=1.$$

3. $$\log_{\frac12}\left(2\sin\frac{\pi}{12}\right)+\log_{\frac12}\left(\cos\frac{\pi}{12}\right)$$

   $$=\log_{\frac12}\left(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\right)=\log_{\frac12}\left(\sin\frac{\pi}{6}\right)$$

   $$=\log_{\frac12}\left(\frac12\right)=1.$$

4. $$\log_{\frac{\sqrt3}{2}}\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\frac{\sqrt3}{2}}\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$

   $$=\log_{\frac{\sqrt3}{2}}\left(\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)\right)$$

   $$=\log_{\frac{\sqrt3}{2}}\left(\cos^2\frac{\pi}{12}-\sin^2\frac{\pi}{12}\right)=\log_{\frac{\sqrt3}{2}}\left(\cos\frac{\pi}{6}\right)$$

   $$=\log_{\frac{\sqrt3}{2}}\left(\frac{\sqrt3}{2}\right)=1.$$

Ответ: а) $$-1$$; б) $$1$$; в) $$1$$; г) $$1$$.