$(\frac{1}{4})^{|2x-1|} > (\frac{4}{5})^{3x}$
$5^{3x} > 4^{3x+|2x-1|}$
กรณี $2x - 1 > 0$ $(x > 0.5)$
$5^{3x} > 4^{5x-1}$
$(3x) log 5 > (5x-1) log 4$
$\frac{log 4}{5 log 4 - 3 log 5} > x$
$\therefore \frac{log 4}{5 log 4 - 3 log 5} > x > 0.5$
กรณี $2x - 1 < 0$ $(x < 0.5)$
$5^{3x} > 4^{x+1}$
$(3x) log 5 > (x+1) log 4$
$x > \frac{log 4}{3 log 5 - log 4}$
$\therefore 0.5 > x > \frac{log 4}{3 log 5 - log 4}$
กรณี $2x - 1 = 0$ $(x = 0.5)$
$5^{3x} > 4^{3x}$
$\therefore x = 0.5$
ดังนั้นคำตอบอยู่ในช่วง $(\frac{log 4}{3 log 5 - log 4}, \frac{log 4}{5 log 4 - 3 log 5})$
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