Question

If $x$ satisfies the equation ${\log }_5(x^2+3x-4)={\log }_5\left(\frac{x-1}{x+4}\right)$, then the value of $\left|x\right|$ is

Answer 1

Clearly, $x^2+3x-4>0,\frac{x-1}{x+4}>0$
$(x-1)(x+4)>0,\frac{x-1}{x+4}>0$
$\Rightarrow x<-4orx>1$ ...(1)
Now, ${\log }_5(x^2+3x-4)={\log }_5\left(\frac{x-1}{x+4}\right)$
$\Rightarrow x^2+3x-4=\frac{x-1}{x+4}$
$\Rightarrow {(x+4)}^2=1\Rightarrow x=-5,-3$ ....(2)
from (1) and (2), we get, $x=-5$
$\therefore |x|=5$
Ans: 5