Answer
$x=2$
Work Step by Step
Apply the logarithmic property:
$\log_a(\dfrac{M}{N}) = \log_a M-\log_a N$
and rearrange the given expression to obtain:
$\log_{1/3}[\dfrac{x^2+x}{x^2-x}] =-1$
Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have:
$\log_{1/3} [\dfrac{x+1}{x-1}] =-1$
$\dfrac{x+1}{x-1}=(\dfrac{1}{3})^{-1}$
or, $\dfrac{x+1}{x-1}=3$
or, $3x^2-3x=x^2+x$
or, $2x^2-4x=0$
or, $x=0, 2$
Since, the domain of the variable is $x \gt 0$, we cannot accept the value of $x=0$
Thus, our answer is: $x=2$
