Verified answer

Ответ:

1

Решение:

[tex]5^{2x-1}+2^{2x}=5^{2x}-2^{2x+2}\\\\2^{2x}+2^{2x+2}=5^{2x}-5^{2x-1}\\\\2^{2x}(1+2^2)=5^{2x}(1-5^{-1})\\\\2^{2x}(1+4)=5^{2x}(1-\frac{1}{5})\\\\2^{2x}*5=5^{2x}*\frac{4}{5}\\\\\frac{2^{2x}}{5^{2x}}=\frac{4}{5*5}\\\\(\frac{2^2}{5^2})^x=\frac{4}{25}\\\\(\frac{4}{25})^x =(\frac{4}{25})^1\\\\x=1[/tex]