# Solve: $\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$

Solve: $$\dfrac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$

My try: Conditions identify: $\left\{ \begin{array}{l} x-1>0\\1-\sqrt[3]{2-x}>0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x>1\\ \sqrt[3]{2-x}<1 \end{array} \right.\Leftrightarrow x>1$

$\dfrac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)\\\Leftrightarrow \log_{\frac{1}{2}}\sqrt{x-1}>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)\\\Leftrightarrow \sqrt{x-1}-1+\sqrt[3]{2-x}<0$

I don't know how to solve this Any equation: $\boxed{\sqrt{x-1}-1+\sqrt[3]{2-x}<0},$ please help me solve that into the result, please guide me, thanks.

$$\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$ so $$\log_{\frac{1}{2}}\left(x-1\right)>2\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$ $$\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}[\left(1-\sqrt[3]{2-x}\right)]^2$$ $$\left(x-1\right)<[\left(1-\sqrt[3]{2-x}\right)]^2$$ $$\left(x-1\right)<\left(1-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$ $$0<\left((2-x)-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$ $$0<\left((2-x)^{\frac{3}{3}}-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$
• I thinks your answre is one of the good key for my question, but can you give me the answer for my stuck question that how to solve $\sqrt{x-1}-1+\sqrt[3]{2-x}<0,$ :) Nov 7 '13 at 15:36
• Actually, $\log_{1/2}$ is decreasing. The inequality should be backwards after that. Nov 7 '13 at 15:39