Example 4 The equation $k \cdot 9^{x}-k \cdot 3^{r+1}+6(k-5)=0$ has a solution for $x \in[0,2]$, find the range of real number $k$.

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Let $t=3^{x}$, since $x \in[0,2]$, then $t \in[1,9]$, the original equation becomes $k t^{2}-3 k t+6(k-$ 5) $=0$. $k=0$ is not valid, thus
$$
t^{2}-3 t+6\left(1-\frac{5}{k}\right)=0 .
$$

The original problem is equivalent to the equation (1) having a solution in $t \in[1,9]$.
Let $f(t)=t^{2}-3 t+6\left(1-\frac{5}{k}\right)$, then
$$
\left\{\begin{array}{l}
\Delta=9-24\left(1-\frac{5}{k}\right) \geqslant 0, \\
f(9)=81-27+6\left(1-\frac{5}{k}\right) \geqslant 0,
\end{array}\right.
$$

Solving this yields $\left[\frac{1}{2}, 8\right]$.