How do you calculate ##antilog 0.3010 ##?

##anti log 0.3010 = 10^(0.3010) ~~ 2##
The function:
##f(x) = 10^x##
is one to one and strictly monotonically increasing on its ##(-oo, oo)##, with ##(0, oo)##
It therefore has an inverse with domain ##(0, oo)## and range ##(-oo, oo)##, namely:
##g(x) = log x##
which is also one to one and strictly monotonically increasing.
Hence the inverse of ##log x## is ##10^x##.
So both of the following hold:
##log(10^x) = x## for all ##x in (-oo, oo)##
##10^log(x) = x## for all ##x in (0, oo)##
##color(white)()##
So:
##anti log 0.3010 = 10^0.3010 ~~ 1.9998618696##
Alternatively, if you know:
##log 2 ~~ 0.30102999566##
then:
##2 ~~ anti log 0.30102999566 ~~ anti log 0.3010##