2^{2x+1} = 5*4^{x-2} + 3^{2x-1}
2^{2x+1} = 5*2^ (2(x-2)) + 3^{2x-1}
2^{2x+1} = 5*2^ (2x-4 )) + 3^{2x-1}
2^{2x+1} - 5*2^ (2x-4 )) = 3^{2x-1}
2^5 *2^{2x - 4} - 5*2^ (2x-4 )) = 3^{2x-1}
32 *2^{2x - 4} - 5*2^ (2x-4 )) = 3^{2x-1}
27 * 2^{2x - 4} =3^3* 3^{2x - 4} | : 27
2^{2x-4} = 3^{2x-4}
Exponenten sind gleich, Basen verschieden.
==> beide Exponenten müssen 0 sein.
2x -4 = 0
==> x = 2.
Probe:
2^{2*2+1} = 5*4^{2-2} + 3^{2*2-1} ?
2^5 = 32
5*4^{0} + 3^{3} = 5 + 27 = 32.
Fortsetzung deiner Rechnung:
27/16 = 3^{2x - 1}/2^{2x}
3^3/2^4 = 3^{2x} / ((2^{2x} * 3)
3^4 / 2^4 = 3^{2x}/(2^{2x})
(3/2)^4 = (3/2)^{2x} | Exponentenvergleich
4 = 2x
==> x = 2.
Zumindest dein Zwischenresultat war also richtig.