How do you calculate #antilog 0.3010 #?

Question

13059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-19T07:51:17

#anti log 0.3010 = 10^(0.3010) ~~ 2#

The function:

#f(x) = 10^x#

is one to one and strictly monotonically increasing on its domain #(-oo, oo)#, with range #(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#