Решить уравнение $$log_8 x + log_{\sqrt{2}} x = 14$$.
Свойство логарифма: $$log_a x = \frac{log_b x}{log_b a}$$
$$log_8 x = \frac{log_2 x}{log_2 8} = \frac{log_2 x}{3}$$
$$log_{\sqrt{2}} x = \frac{log_2 x}{log_2 \sqrt{2}} = \frac{log_2 x}{\frac{1}{2}} = 2log_2 x$$
$$\frac{log_2 x}{3} + 2log_2 x = 14$$
$$log_2 x (\frac{1}{3} + 2) = 14$$
$$log_2 x \cdot \frac{7}{3} = 14$$
$$log_2 x = 14 \cdot \frac{3}{7}$$
$$log_2 x = 2 \cdot 3$$
$$log_2 x = 6$$
По определению логарифма: $$log_a b = c$$ $$a^c = b$$
$$x = 2^6$$
$$x = 64$$
Проверим корень: $$log_8 64 + log_{\sqrt{2}} 64 = log_8 8^2 + log_{\sqrt{2}} (\sqrt{2})^{12} = 2 + 12 = 14$$
Ответ: x = 64