China Mathematical Competition

$|\lg (a+1)|=|\lg \left(-\frac{b+1}{b+2}+1\right)|=|\lg \left(\frac{1}{b+2}\right)|=|\lg (b+2)|.$ Then either $a+1=b+2$ or $(a+1)(b+2)=1$. Since $a<b$, so $a+1\ne b+2$. Therefore, $(a+1)(b+2)=1$. From $f(a)=|\lg (a+1)|$ we know $0<a+1<1$. Then $$0<a+1<b+1<b+2,$$ which implies $$0<a+1<1<b+2.$$ Therefore, $$\begin{gathered} (10a+6b+21)+1=10(a+1)+6(b+2) \\ =6(b+2)+\frac{10}{b+2}>1. \end{gathered}$$ Then $$\begin{gathered} f(10a+6b+21)=\left|\lg \left[6(b+2)+\frac{10}{b+2}\right]\right| \\ =\lg \left[6(b+2)+\frac{10}{b+2}\right]. \end{gathered}$$ On the other hand, $$f(10a+6b+21)=4\lg 2.$$ So $$\lg \left[6(b+2)+\frac{10}{b+2}\right]=4\lg 2,$$ which means $6(b+2)+\frac{10}{b+2}=16$. Then either $b=-\frac{1}{3}$ or $b=-\frac{2}{5}$ (discarded). Substituting $b=-\frac{1}{3}$ into $(a+1)(b+2)=1$, we find $a=-\frac{2}{5}$. Therefore $a=-\frac{2}{5}$, $b=-\frac{1}{3}$. $\square$