# How do you calculate antilog 0.3010 ?

Jun 19, 2016

$a n t i \log 0.3010 = {10}^{0.3010} \approx 2$

#### Explanation:

The function:

$f \left(x\right) = {10}^{x}$

is one to one and strictly monotonically increasing on its domain $\left(- \infty , \infty\right)$, with range $\left(0 , \infty\right)$

It therefore has an inverse with domain $\left(0 , \infty\right)$ and range $\left(- \infty , \infty\right)$, namely:

$g \left(x\right) = \log x$

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:

$\log \left({10}^{x}\right) = x$ for all $x \in \left(- \infty , \infty\right)$

${10}^{\log} \left(x\right) = x$ for all $x \in \left(0 , \infty\right)$

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So:

$a n t i \log 0.3010 = {10}^{0.3010} \approx 1.9998618696$

Alternatively, if you know:

$\log 2 \approx 0.30102999566$

then:

$2 \approx a n t i \log 0.30102999566 \approx a n t i \log 0.3010$