$2\log_{6}6 + 3\log_{2}8 - 4\log_{3}1$

$\frac{-3x+5}{3} = 27^{2x-1}$

$\frac{2}{3}\log_{10}1 + 5\log_{\sqrt{3}}3 - 4\log_{\sqrt{2}}1 + \frac{7}{2}\log_{6}6 = $

$\log_{3}(3x-2) = 5$

$\log_{2}8 = x+1$

$2^{5x+1} = 8$

$(\frac{3}{4})^{x} = 16$

$x^{-5} = 32$

$3\log_{5}125 = x$

$2^{3x+2} = 64^{-2}$