1/3 log_{1/2}^2 x + log_{1/2} x + 2log_{1/2} 24 - 2log_{1/2}^3 = ?

Решим данное выражение:

$$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2log_{\frac{1}{2}} 24 - 2log_{\frac{1}{2}} 3 =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 (log_{\frac{1}{2}} 24 - log_{\frac{1}{2}} 3) =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 log_{\frac{1}{2}} (24/3) =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 log_{\frac{1}{2}} 8 =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 log_{\frac{1}{2}} (\frac{1}{2})^{-3} =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 \cdot (-3) log_{\frac{1}{2}} (\frac{1}{2}) =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x + 2 \cdot (-3) \cdot 1 =$$ $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x - 6$$

Ответ: $$1/3 log_{\frac{1}{2}}^2 x + log_{\frac{1}{2}} x - 6$$