Answer:
[tex]I(db)=320[/tex]
Step-by-step explanation:
Given
[tex]I(db)=10\log(\frac{I}{I_o})[/tex]
[tex]I = 10^{32}(I_o)[/tex]
Required
Determine [tex]I(dB)[/tex]
We have:
[tex]I(db)=10\log(\frac{I}{I_o})[/tex]
Substitute [tex]I = 10^8(I_o)[/tex]
[tex]I(db)=10\log(\frac{10^{32}(I_o)}{I_o})[/tex]
[tex]I(db)=10\log(10^{32})[/tex]
Apply law of logarithm
[tex]I(db)=10*32\log(10)[/tex]
[tex]\log(10) = 1[/tex]
So:
[tex]I(db)=10*32*1[/tex]
[tex]I(db)=320[/tex]