Solution
= Solution
Use the same <general solution>. The first two conditions now give $A+B=3$ and $-2A-6B+12=-2$, so $A=1$, $B=2$. Hence
$$
\boxed{f(t)=e^{-2t}-3e^{-4t}+2e^{-6t}}.
$$
At $t=\log\sqrt2$, its value is $1/2-3/4+2/8=0$, so the third condition is also satisfied. This is the unique consistent solution.