Question

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?

Answer 1

I = 10 log ( I / I₀)

23 = 10  log ( I₁ / I₀)

61 =  10 log ( I₂ / I₀)

Subtracting we get

61 - 23 = 10 ( log ( I₂/I₁) )

3.8 = log (I₂/I₁)

I₂/I₁ = 10^(3.8)

I_{2 =} 6309.57I₁

Answer 2

$B = 10(dB) log(\frac{I}{10^{-12}})$

$Given, B_{i} = 23dB , B_{f} = 62 dB$

$B_{i} = 10(dB) log(\frac{I_{i}}{10^{-12}}) = 23 dB$

$B_{f} = 10(dB) log(\frac{I_{f}}{10^{-12}}) = 61 dB$

Subtracting both equation

$10(dB) log(\frac{I_{f}}{10^{-12}}) - 10dB log(\frac{t_{i}}{10^{-12}}) = 61 dB - 23 dB$

$10(dB) log(\frac{I_{f}}{I_{i}})= 38 dB$

$log (\frac{I_{f}}{I_{i}}) = 3.8$.

$\frac{I_{f}}{I_{i}} = 10^{3.8}$

$\frac{I_{f}}{I_{i}} = 6309.57$

Answer 3

I = 10 log ( I / I₀)

23 = 10  log ( I₁ / I₀)

61 =  10 log ( I₂ / I₀)

Subtracting we get

61 - 23 = 10 ( log ( I₂/I₁) )

3.8 = log (I₂/I₁)

I₂/I₁ = 10^(3.8)

I₂/I₁ = 6309.57

I_{2 =} 6309.57 I₁