Question

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

Answer 1

$log_5(2x-1)+log_5(4-x)=1\O.D.3.:\begin{cases}2x-1 textgreater 0\4-x textgreater 0end{cases}Rightarrowbegin{cases}x textgreater frac12\x textless 4end{cases}Rightarrow xin(frac12;;4)\log_5((2x-1)(4-x))=1\9x-2x^2-4=5^1\2x^2-9x+9=0\D=81-4cdot2cdot9=9\x_{1,2}=frac{9pm3}4\x_1=3,;x_2=1,5$