Log_5(2x-1)+log_5(4-x)=1

[latex]\log_5(2x-1)+\log_5(4-x)=1\\O.D.3.:\\\begin{cases}2x-1\ \textgreater \ 0\\4-x\ \textgreater \ 0\end{cases}\Rightarrow\begin{cases}x\ \textgreater \ \frac12\\x\ \textless \ 4\end{cases}\Rightarrow x\in(\frac12;\;4)\\\log_5((2x-1)(4-x))=1\\9x-2x^2-4=5^1\\2x^2-9x+9=0\\D=81-4\cdot2\cdot9=9\\x_{1,2}=\frac{9\pm3}4\\x_1=3,\;x_2=1,5[/latex]