The decibel level D of a sound is related to its intensity I by D=10log(I/I_0). If I_0 is 10^-12, then what is the intensity of a noise measured at 48 decibels? Express your answer in scientific notation, rounding to three significant digits, if necessary?

Question

danjs · Answer

so solve

$$48 = 10*\log(\frac{ I }{ 10^{-12} })$$

thatonegirl_ · Answer

Yes

danjs · Answer

$$\frac{ 48 }{ 10 } = \log \frac{ I }{ 10^{-12} }$$

danjs · Answer

$$10^{\frac{ 48 }{ 10 }} = \frac{ I }{ 10^{-12} }$$

thatonegirl_ · Answer

why does it turn to 10 when you divide by the log?

danjs · Answer

I =  10^(-12) * 10^(48/10)

danjs · Answer

no, recall ...

danjs · Answer

$$10^{\log(x)} = x$$

thatonegirl_ · Answer

ohhh nvm sorry continue

thatonegirl_ · Answer

u just rewrote it in exponential form, right?

danjs · Answer

to solve for I, you need to get rid of the log

danjs · Answer

by doing 10^(everything) on both sides

danjs · Answer

where 10 is the base of the log

thatonegirl_ · Answer

okay i see

danjs · Answer

so you just have to calculate 

I = 10^(-12) * 10^(48/10)

thatonegirl_ · Answer

so it comes out to be 6.31x10^-8 watt/m^2 ?

danjs · Answer

yeah, that is what i just got too

danjs · Answer

is that a reasonable range for the Intensity? I can't recall

thatonegirl_ · Answer

okay cool, thanks a ton for all ur help!

thatonegirl_ · Answer

idk lol its an answer choice. i plugged it back into the original equation and it works.

danjs · Answer

ha, k cool

thatonegirl_ · Answer

thanks :D

danjs · Answer

np