How do you calculate antilog 0.3010 ?

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##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)##, delay ##(0, oo)##

It consequently has an inverse delay territory ##(0, oo)## and stroll ##(-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 twain of the forthcoming 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##

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