Example 3 In a sequence of positive numbers, $a_{1}=10, a_{k-1}=10 \sqrt{a_{k}}$, find the general formula for $a_{n}$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Analysis: For a recursive sequence of the form $a_{n+1}=a a_{n}^{p} a_{n-1}^{q}$, the general term can be solved by taking the logarithm on both sides, transforming it into a second-order recursive sequence problem.
Solution: It is easy to see that for any positive integer $k, a_{k}>0$. Taking the logarithm on both sides of the recursive formula, we get $\lg a_{k \cdot 1}=1+\frac{1}{2} \lg a_{k}$. Let $\lg a_{k}=b_{k}$, then $b_{k+1}=\frac{1}{2} b_{k}+1$, and $b_{1}=1, b_{2}=\frac{3}{2}$.
We can assume $b_{k}=A \cdot\left(\frac{1}{2}\right)^{k}+B$, then $\left\{\begin{array}{l}b_{1}=\frac{A}{2}+B \\ b_{2}=\frac{A}{4}+B\end{array}\right.$, leading to $\left\{\begin{array}{l}A=-2 \\ B=2\end{array}\right.$. Therefore, $b_{k}=-\left(\frac{1}{2}\right)^{k-1}+2$.
Since $\lg a_{k}=b_{k}$, we have $a_{k}=10^{-\left(\frac{1}{2}\right)^{k}}+2(k \in \mathbf{N})$.