7. If the equation $5^{x}=\frac{a+3}{5-a}$ has a negative root for $x$, then the range of real values for $a$ is $\qquad$ .
7. $-3<a<1 . x<0$, then $5^{x} \in(0,1)$, i.e., $0<\frac{a+3}{5-a}<1$, which can be solved to get $-3<a<1$.