For a B-tree with minimum degree $t$, every internal node except root has at least $t$ children.

Also, every node except root has at least $t-1$ keys.

For a B-tree of height $h$, the minimum number of keys is:

$n\ge 2t^h-1$

Now solve this inequality for $h$.

$n+1\ge 2t^h$

$\frac{n+1}{2}\ge t^h$

Taking logarithm base $t$ on both sides:

$\log_t\frac{n+1}{2}\ge h$

So,

$h\le \log_t\frac{n+1}{2}$