Encoding huge number of tape-symbols of a turing machine in the simulation of the turing-machine using real computer I was going through the classic text "Introduction to Automata Theory, Languages and Computation" by Hopcroft,Ullman,Motwani where I came across the simulation of a turing machine using a real computer. There the author argued that it shall be almost impossible to carryout the above simulation (not considering universal turing machine) if the number of tape symbols are quite huge. In such a situation it might happen that the code of a tape symbol might not fit in the single hard-disk of a computer.

Then the author makes a claim as:


  There would have to be very many tape symbols indeed, since a 30 gigabyte disk, for instance, can represent any of $2^{240000000000}$ symbols.


Now I can't figure out the specific mathematics that the author does...

I assume the usual encoding as we do in digital logic say to encode $8$ symbols we need atleast $3$ bits of code.
Then to represent $n$ symbols if $k$ bits are required then we should have the following relation,

$2^{k} = n$ 

$\implies$ $k=log_2(n)$

Now,

$30$ $gigabytes$ $= 30 \times 2^{33}$ $bits$ $= \beta (say)$

Now if our disk can hold all the $n$ symbols then we shall have the following relation true:

$Disk$ $size$ $= (Bits$ $required$ $for$ $each$ $symbol$) $\times$ ($number$ $of$ $symbols)$

$\implies$ $\beta = k \times n = (log_2(n)\times n)$

Solving graphically I have:

 

$n= 7.8 \times 10^{9}$ , which is no where close to the number $2^{240000000000}$

Where am I making the mistake?
You are interpreting this statement backwards - he is not calculating how many symbols can 30 gigabytes represent, he is calculating how many symbols can be represented by using 30 gigabytes to represent a single symbol!  That would be necessary to be able to approximate a calculation where each symbol is an infinitely-precise real number.