a)
[tex](\frac{1}{2})^{3x-4}=8^{2x+1}\\\\(\frac{1}{2})^{3x-4}=((\frac{1}{2})^{-3})^{2x+1}\\\\(\frac{1}{2})^{3x-4}=(\frac{1}{2})^{-6x-3}\\\\3x-4=-6x-3\\\\3x+6x=-3+4\\\\9x=1 \ \ /:9\\\\\boxed{x=\frac{1}{9}}[/tex]
b)
[tex](\frac{3}{5})^{x+2}\cdot(\frac{25}{9})^{-2x+3}=(0,6)^{x^2}\\\\(\frac{3}{5})^{x+2}\cdot((\frac{5}{3})^2)^{-2x+3}=(\frac{3}{5})^{x^2}\\\\(\frac{3}{5})^{x+2}\cdot((\frac{3}{5})^{-2})^{-2x+3}=(\frac{3}{5})^{x^2}\\\\(\frac{3}{5})^{x+2}\cdot(\frac{3}{5})^{4x-6}=(\frac{3}{5})^{x^2}\\\\(\frac{3}{5})^{x+2+4x-6}=(\frac{3}{5})^{x^2}\\\\(\frac{3}{5})^{5x-4}=(\frac{3}{5})^{x^2}\\\\5x-4=x^2\\\\-x^2+5x-4=0 \ \ /\cdot(-1)\\\\x^2-5x+4=0\\\\a=1, \ b=-5, \ c=4\\\\\Delta=(-5)^2-4\cdot1\cdot4=25-16=9\\\\\sqrt{\Delta}=\sqrt9=3[/tex]
[tex]x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2\cdot1}=\frac{5-3}{2}=\frac{2}{2}=\boxed1\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2\cdot1}=\frac{5+3}{2}=\frac{8}{2}=\boxed4[/tex]
