Log(x)=4.203
10^4.203=x
10^4 * 10^.203 =x
sooo, 10^.203 is a wee bit less than 10^.33333, which, is turn, is a wee bit more than 2. SO lets estimate 10^.203 as "about 2"
Which gives us "about 20000"
which is accurate to one signifigant figure, which is about as good as it will be with now machine-assistance. Or log tables.
I suppose you could work out a taylor series for this and calculate X, using 20000 as the starting point for x to keep accuracy up. It would be tedious, but doable. That said, I'm not going to do it.