How do you calculate $antilog 0.3010$ ?

Question

14846 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:

$2 ~~ anti log 0.30102999566 ~~ anti log 0.3010$

How do you calculate antilog 0.3010 antilog 0.3010 ?