[tex]\bf log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-----------------------------\\\\
2^{x+1}=5^{2x-1}\implies log(2^{x+1})=log(5^{2x-1})
\\\\\\
(x+1)log(2)=(2x-1)log(5)\implies (x+1)\cfrac{log(2)}{log(5)}=(2x-1)
\\\\\\
now\quad \cfrac{log(2)}{log(5)}\textit{ is just a constant about }0.43068[/tex]
and you can simply solve that for "x"
