Question

please help me I'm desperate.

Answer 1

$D:4x-1>0 \wedge x+2 >0 \wedge x>0\\ D:4x>1 \wedge x>-2 \wedge x>0\\ D:x>\frac{1}{4} \wedge x>0\\ D:x>\frac{1}{4}\\\\ \log(4x-1)-\log(x+2)=\log x\\ \log\frac{4x-1}{x+2}=\log x\\ \frac{4x-1}{x+2}=x\\ x(x+2)=4x-1\\ x^2+2x=4x-1\\ x^2-2x+1=0\\ (x-1)^2=0\\ \boxed{x=1}$

Answer 2

$\log (4x - 1) - \log (x + 2) = \log x$
$\log ( \frac{4x - 1}{x + 2} ) = \log x$
$\frac{4x - 1}{x + 2} = x$
$4x - 1 = x(x + 2)$
$4x - 1 = x^2 + 2x$
$0 = x^2 - 2x + 1$
$0 = (x - 1)^2$
⇒ $x = 1$