1\; \; \; \Rightarrow \\\\2log_25>3log_{\frac{1}{8}}\frac{1}{24}\\\\2)\; \; log_{b}a=\sqrt3 " alt=" 1)\; \; 3\cdot log_{\frac{1}{8}}\frac{1}{24}=3\cdot log_{2^{-3}}\frac{1}{24}=-3\cdot \frac{1}{3}\cdot log_2\, 24^{-1}=log_2\, 24\\\\2log_25=log_225\\\\log_2241\; \; \; \Rightarrow \\\\2log_25>3log_{\frac{1}{8}}\frac{1}{24}\\\\2)\; \; log_{b}a=\sqrt3 " align="absmiddle" class="latex-formula">![log_{\frac{\sqrt{a}}{b}}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{log_{b}\frac{\sqrt[3]{a}}{\sqrt{b}}}{log_{b}\frac{\sqrt{a}}{b}}=\frac{log_{b}\sqrt[3]{a}-log_{b}\sqrt{b}}{log_{b}\sqrt{a}-log_{b}b}=\frac{\frac{1}{3}log_{b}a-\frac{1}{2}log_{b}b}{\frac{1}{2}log_{b}a-1}=\\\\=\frac{\frac{1}{3}\cdot \sqrt3-\frac{1}{2}}{\frac{1}{2}\cdot \sqrt3-1}=\frac{(2\sqrt3-3)\cdot 2}{6\cdot (\sqrt3-2)}=\frac{\sqrt3\cdot (2-\sqrt3)}{3\cdot (\sqrt3-2)}=-\frac{\sqrt3}{3}=-\frac{1}{\sqrt3}](https://tex.z-dn.net/?f=+log_%7B%5Cfrac%7B%5Csqrt%7Ba%7D%7D%7Bb%7D%7D%5Cfrac%7B%5Csqrt%5B3%5D%7Ba%7D%7D%7B%5Csqrt%7Bb%7D%7D%3D%5Cfrac%7Blog_%7Bb%7D%5Cfrac%7B%5Csqrt%5B3%5D%7Ba%7D%7D%7B%5Csqrt%7Bb%7D%7D%7D%7Blog_%7Bb%7D%5Cfrac%7B%5Csqrt%7Ba%7D%7D%7Bb%7D%7D%3D%5Cfrac%7Blog_%7Bb%7D%5Csqrt%5B3%5D%7Ba%7D-log_%7Bb%7D%5Csqrt%7Bb%7D%7D%7Blog_%7Bb%7D%5Csqrt%7Ba%7D-log_%7Bb%7Db%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7Dlog_%7Bb%7Da-%5Cfrac%7B1%7D%7B2%7Dlog_%7Bb%7Db%7D%7B%5Cfrac%7B1%7D%7B2%7Dlog_%7Bb%7Da-1%7D%3D%5C%5C%5C%5C%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%5Ccdot+%5Csqrt3-%5Cfrac%7B1%7D%7B2%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot+%5Csqrt3-1%7D%3D%5Cfrac%7B%282%5Csqrt3-3%29%5Ccdot+2%7D%7B6%5Ccdot+%28%5Csqrt3-2%29%7D%3D%5Cfrac%7B%5Csqrt3%5Ccdot+%282-%5Csqrt3%29%7D%7B3%5Ccdot+%28%5Csqrt3-2%29%7D%3D-%5Cfrac%7B%5Csqrt3%7D%7B3%7D%3D-%5Cfrac%7B1%7D%7B%5Csqrt3%7D+ "log_{\frac{\sqrt{a}}{b}}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{log_{b}\frac{\sqrt[3]{a}}{\sqrt{b}}}{log_{b}\frac{\sqrt{a}}{b}}=\frac{log_{b}\sqrt[3]{a}-log_{b}\sqrt{b}}{log_{b}\sqrt{a}-log_{b}b}=\frac{\frac{1}{3}log_{b}a-\frac{1}{2}log_{b}b}{\frac{1}{2}log_{b}a-1}=\\\\=\frac{\frac{1}{3}\cdot \sqrt3-\frac{1}{2}}{\frac{1}{2}\cdot \sqrt3-1}=\frac{(2\sqrt3-3)\cdot 2}{6\cdot (\sqrt3-2)}=\frac{\sqrt3\cdot (2-\sqrt3)}{3\cdot (\sqrt3-2)}=-\frac{\sqrt3}{3}=-\frac{1}{\sqrt3}")