11. If the equation $\lg \left(4 x^{2}+4 a x\right)-\lg (4 x-a+1)=0$ has a unique real solution, find the range of values for $a$.
11. From $4 x-a+1>0$, we get $x>\frac{a-1}{4}$. From the original equation, we know $4 x^{2}+4 a x=4 x-a+1>0$, which is $4 x^{2}+4(a-1) x+a-1=0, \Delta=16(a-1)^{2}-16(a-1) \geqslant 0$, yielding $a \leqslant 1$ or $a \geqslant 2$. When $a=1$ or $a=2$, the original equation has no solution; when $a>2$, the above quadratic equation has two distinct real roots $\left(x_{1}<\frac{a-1}{4}, x_{1} \leqslant \frac{a-1}{4}\right)$. Let $f(x)=4 x^{2}+4(a-1) x+a-1$, then $f\left(\frac{a-1}{4}\right) \leqslant 0$, yielding $\frac{1}{5} \leqslant a \leqslant 1$. When $a=\frac{1}{5}$, $x_{1}=1, x_{2}=-\frac{1}{5}$ (discard). In summary, when $\frac{1}{5} \leqslant a<1$, the original equation has a unique solution.