[tex]a)\\\log_{2\sqrt2}4=\log_{2^\frac32}2^2=\log_{2^\frac32}\left[\left(2^\frac32\right)^\frac23\right]^2=\log_{2^\frac32}\left(2^\frac32\right)^\frac43=\frac43\cdot\log_{2^\frac32}2^\frac32=\frac43\\\\\\b)\\\log_{25}5\sqrt5=\log_{5^2}5^\frac32=\log_{5^2}\left[\left(5^2\right)^\frac12\right]^\frac32=\log_{5^2}\left(5^2\right)^\frac34=\frac34\log_{5^2}5^2=\frac34[/tex]
[tex]c)\\\log_{0,01}\dfrac{\sqrt[3]{100}}{10} =\log_{10^{-2}}\dfrac{\sqrt[3]{10^2}}{10} =\log_{10^{-2}}\dfrac{10^\frac23}{10^1} =\log_{10^{-2}}10^{-\frac13} =\\\\=\log_{10^{-2}}\left[\left(10^{-2}\right)^{-\frac12}\right]^{-\frac13} =\log_{10^{-2}}\left(10^{-2}\right)^{\frac16} =\frac16\log_{10^{-2}}10^{-2}=\frac16 \\\\\\ d)\\\log_{\frac16}\dfrac{\sqrt6}{36} =\log_{6^{-1}}\dfrac{6^\frac12}{6^2} =\log_{6^{-1}}\big6^{-1\frac12}=\log_{6^{-1}}\big(6^{-1}\big)^{1\frac12}=1\frac12\log_{6^{-1}}\big6^{-1}=1\frac12[/tex]
