How does augmenting linear bounded automata tape alphabets increase memory?
If you consider an input, hence a tape, of size $n$ and a tape alphabet of
$b$ symbols, you get $b^n$ configurations possible on your tape. If
you measure memory in bits, the memory size is the base 2 logarithm of
the number of configurations, very simply because that is the number
of bits required to have the same number of configurations (rounding up
to the next integer, because half-bits have not been invented yet).

So the tape memory of your linear bounded automaton (LBA) is $\log_2
b^n$, which is equal to $n\log_2 b$.

Augmenting your tape alphabet does not increase your memory very
fast. Doubling the size of the alphabet is equivalent to a single extra bit per cell, i.e. adding $n$ bits total. The
positive side of it is that it does not take many bits per cell of the
tape to have a huge alphabet.