Hello, kevin3000!
This is a strange problem . . .
Calculate an approximate value to $\displaystyle 128^{48}$
if you know that: $\displaystyle 2^{14} = 16,\!384$ . . I don't see how this is useful.
I would do it like this . . .
We have: .$\displaystyle N \;=\;128^{48} \:=\:\left(2^7\right)^{48} \:=\:2^{336} $
Since $\displaystyle 2^{10} \:=\:1024 \:\approx\:1000 \:=\:10^3$
we have: .$\displaystyle N \;=\;2^6\cdot2^{330} \;=\;2^6\cdot (2^{10})^{33}$
. . . . . . . .$\displaystyle N\;\approx\;2^6\cdot(10^3)^{33} \;=\;64\cdot10^{99}$
Therefore: .$\displaystyle N \;\approx\;6.4 \times 10^{100}$
It's funny yes but I was merely following the instructions of the problem (or trying to do so, since the problem doesn't make much sense); given freedom, I would also use the approximation 2^10 $\displaystyle \approx$ 10^3.
Anyway if you go high enough, 2^10 $\displaystyle \approx$ 10^3 will give you 99.99% error as well.
[QUOTE=1005;539205]undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)
and Soroban's approximation is 1 order of magnitude off with 54.28% error.[/QUOTE
Soroban's approx is not one magnitude off. It can be expressed as 0.64E101. Calculator answer is 1.4xE 101
bjh