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Answer 1

$displaystyle left({3over7}right)^{3x+1}=left({7over3}right)^{5x-3}\\ left({3over7}right)^{3x+1}=left({3over7}right)^{3-5x}\\ 3x+1=3-5x\ 8x=2\\ mathbf{x={1over4}}$

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$displaystyle log_{pi}{(x^2+2x+3)}=0\\ x^2+2x+3>0\ D=4-12=-8<0,;; xinmathbb{R}\\ log_{pi}{(x^2+2x+3)}=log_{pi}{1}\ x^2+2x+3=1\ x^2+2x+2=0\ D=4-8=-4<0,;; x_{1},x_{2}notinmathbb{R}\ sqrt{D}=2i\\ x_{1}={-2+2iover2}=-1+i\x_{2}={-2+2iover2}=-1-i\\ x_{1},x_{2}inmathbb{C}$

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$displaystyle log_{3}{(4-2x)}-log_{3}{2}=2\\ 4-2x>0\ x<2\\ log_{3}{4-2xover2}=log_{3}9\\ 4-2x=18\2x=-14\ mathbf{x=-7}$