How Does a 2.80 dB Difference Affect Sound Intensity?

[Oijl]

Homework Statement

Two sounds differ in sound level by 2.80 dB. By what factor is the one intensity greater than the other?

Homework Equations

The Attempt at a Solution

Is the equation

B = 10dB log(I/Io)

relevant?

 

[rl.bhat]

The equation is
Bdb = 10log(I/Io)
Now let B1 = 10log(I1/Io) and
B2 = 10log(I2/Io)
Take the difference and find the ratio of I1/I2.

 

[danago]

Lets say sound 1 has an intensity of I₁ and intensity level B, and sound 2 has an intensity of I₂ with intensity level B+2.8

B = 10 log (I₁ / I₀)
B+2.8 = 10 log (I₂ / I₀)

See if you can use those

EDIT: rl.bhat beat me to it :P

 

[Oijl]

I'm sorry, but the algebra is confusing me. Could you show the first few steps, please?

 

[rl.bhat]

B1 = 10log(I1/Io)
B2 = 10log(I2/Io)
B1 -B2 = 10[log(I1/Io) - log(I2/Io)]
Use laws of logarithm to further simplification.