$\displaystyle B=log(\frac{I}{I_{0}})$
If B=58.4 dB and $\displaystyle I_{0}=10^{-12} \frac{W}{m^2}$
then:
$\displaystyle 58.4=log(\frac{I}{10^{-12}})$
log(I/
If you are wanting to know $\displaystyle I $ it is as follows:
$\displaystyle 58.5=log(\frac{I}{10^{-12}}) $
$\displaystyle 10^{58.5}=\frac{I}{10^{-12}} $
$\displaystyle (10^{58.5})(10^{-12})=I $
$\displaystyle I=3.16 \times 10^{46} $