25^(x-1) - 129 * 5^(x-2) + 20<0

$25^{x-1} -129* 5^{x-2} +20\ \textless \ 0$
$\frac{ 25^{x} }{25 ^{1} }-129* \frac{ 5^{x} }{ 5^{2} } +20\ \textgreater \ 0 |*25 ( 5^{x} ) ^{2} -129* 5^{x} +500\ \textgreater \ 0$
$5^{x}=t, t\ \textgreater \ 0$
t²-129t+500>0
t₁=4, t₂=125
+ - +
------------(4)---------(125)----------------->t

t<4, t>125
$1. 5^{x} \ \textless \ 4, log_{5} 5^{x} \ \textless \ log_{5} 4$
$x\ \textless \ log_{5} 4$
$2. t\ \textgreater \ 125 5^{x} \ \textgreater \ 125 5^{x} \ \textgreater \ 5^{3}$
a=5, 5>1. знак неравенства не меняем
x>3
ответ: x∈(-∞; log₅4)∪(3;∞)