What is the relationship between sound intensity level and power?

[LeakyFrog]

Homework Statement

What fraction of the acoustic power of a noise would have to be eliminated to lower its sound intensity level from 90 to 70dB?

Homework Equations

I = Power/Area
B = 10 * log *(I/I₀)

I₀ = 10^-12W/m²

The Attempt at a Solution

The answer I got for this is that the new power is 1% of the original. Which I don't think is true. A nudge in the right direction would be appreciated

 

[willem2]

You can just solve $B = 10 * log (I/I_0)$ for I with B=90 and B is 70

remember to use the base 10 logarithm and not the natural logarithm

 

[LeakyFrog]

You can just solve $B = 10 * log (I/I_0)$ for I with B=90 and B is 70

remember to use the base 10 logarithm and not the natural logarithm

Thanks. I guess it is 1% of the original power after all. Seems kind of strange that it would have to drop so dramatically for only 20dB.