Question 122, In the sequences $\left\{a_{n}\right\}, \left\{b_{n}\right\}$, $a_{1}=b_{1}=10, \left\{\begin{array}{l}a_{n+1}=\frac{1}{a_{n} b_{n}} \\ \mathrm{~b}_{\mathrm{n}+1}=a_{n}^{4} \mathrm{~b}_{\mathrm{n}}\end{array}\left(\mathrm{n} \in \mathrm{Z}^{+}\right)\right.$, find $\mathrm{a}_{\mathrm{n}}, \mathrm{~b}_{\mathrm{n}}$.
Question 122, Solution: Taking the logarithm, we have $\left\{\begin{array}{l}\lg a_{n+1}=-3 \lg a_{n}-\lg b_{n} \\ \lg b_{n+1}=4 \lg a_{n}+\lg b_{n}\end{array}\right.$ . Let $x_{n}=\lg a_{n}, y_{n}=\lg b_{n}$, then $\begin{aligned} x_{1}= & y_{1}=1, \text { and } \\ & \left\{\begin{array}{c}x_{n+1}=-3 x_{n}-y_{n} \\ y_{n+1}=4 x_{n}+y_{n}\end{array}\right.\end{aligned}$
Adding the two equations gives $x_{n+1}+y_{n+1}=x_{n} \Rightarrow y_{n+1}=x_{n}-x_{n+1}$ . Therefore,
$$
x_{n+2}=-3 x_{n+1}-y_{n+1}=-3 x_{n+1}-\left(x_{n}-x_{n+1}\right)=-2 x_{n+1}-x_{n}
$$

The characteristic equation is $x^{2}+2 x+1=0$, which has a double real root -1, so there exist constants $A, B$ such that $x_{n}=(A+$ $Bn)(-1)^{n}$.
Given $x_{1}=1, x_{2}=-4$, we have $x_{n}=(2-3 n)(-1)^{n}$. Consequently,
$$
y_{n}=x_{n-1}-x_{n}=[2-3(n-1)](-1)^{n-1}-(2-3 n)(-1)^{n}=(6 n-7)(-1)^{n}
$$

Thus, $a_{n}=10^{(2-3 n)(-1)^{n}}, b_{n}=10^{(6 n-7)(-1)^{n}}$.