Logarithm Decibel Problem

[LittlDino]

I have been trying to solve this question for ages now... I know the answer but cannot figure out how they get there.

In acoustics, decibels are used to measure sound pressure according to the law L = 10log₁₀P_{1^{2 - 20log}10}P_0.
Given that P₀ = 2 x 10^-5 and P₁ = 2000 for a rocket launcher, calculate the sound pressure L, created by the rocket launcher.

I have tried to move things around in every possible way and I'm sure I'm just missing something. The answer is 160 (highlight to show) and I can get to 16 but not sure if I only got that because I knew the final answer.

Please help. It's driving me crazy!

 

[Steven G]

I have been trying to solve this question for ages now... I know the answer but cannot figure out how they get there.

In acoustics, decibels are used to measure sound pressure according to the law L = 10log₁₀P_{1^{2 - 20log}10}P_0.
Given that P₀ = 2 x 10^-5 and P₁ = 2000 for a rocket launcher, calculate the sound pressure L, created by the rocket launcher.

I have tried to move things around in every possible way and I'm sure I'm just missing something. The answer is 160 (highlight to show) and I can get to 16 but not sure if I only got that because I knew the final answer.

Please help. It's driving me crazy!

Is the 2 next to P1 the power of P1?

 

[Steven G]

I have been trying to solve this question for ages now... I know the answer but cannot figure out how they get there.

In acoustics, decibels are used to measure sound pressure according to the law L = 10log₁₀P_{1^{2 - 20log}10}P_0.
Given that P₀ = 2 x 10^-5 and P₁ = 2000 for a rocket launcher, calculate the sound pressure L, created by the rocket launcher.

I have tried to move things around in every possible way and I'm sure I'm just missing something. The answer is 160 (highlight to show) and I can get to 16 but not sure if I only got that because I knew the final answer.

Please help. It's driving me crazy!

We need to use log (x/y)=Log x - log y and logx^y= ylog x

L=10logP1^2 - 20logP0 = 20 log P1 - 20 log P0 = 20 (log p1 - log P0) = 20 log (P1/P0)

Let's calculate (P1/P0) = (2* 10^3)/(2*10^-5) =10^8

So 20 log (P1/P0) = 20 Log 10^8 = 160 log 10 = 160 since log 10 =1 and 160 * 1 =160.

 

[Ishuda]

I have been trying to solve this question for ages now... I know the answer but cannot figure out how they get there.

In acoustics, decibels are used to measure sound pressure according to the law L = 10log₁₀P_{1^{2 - 20log}10}P_0.
Given that P₀ = 2 x 10^-5 and P₁ = 2000 for a rocket launcher, calculate the sound pressure L, created by the rocket launcher.

I have tried to move things around in every possible way and I'm sure I'm just missing something. The answer is 160 (highlight to show) and I can get to 16 but not sure if I only got that because I knew the final answer.

Please help. It's driving me crazy!

First of all, as I tell my wife, it's not a drive, just a short walk (at least for some of us).

The decibel (10 bels) is usually reference some level. The equation you have provides the reference level of P₀ = 2 x 10^-5 and we have
L = 10 log₁₀(P_{1 ^/} P₀)² = 20 log₁₀(P_{1 ^/} P₀)
or, if you wish, to the nearest single decimal place,
L = 20 log₁₀(P₁) + 94.0
So, what is 20 log₁₀(2000)?


BTW: There is a different reference for air and underwater acoustics.

 

[LittlDino]

We need to use log (x/y)=Log x - log y and logx^y= ylog x

L=10logP1^2 - 20logP0 = 20 log P1 - 20 log P0 = 20 (log p1 - log P0) = 20 log (P1/P0)

Let's calculate (P1/P0) = (2* 10^3)/(2*10^-5) =10^8

So 20 log (P1/P0) = 20 Log 10^8 = 160 log 10 = 160 since log 10 =1 and 160 * 1 =160.

Thanks soooo much. I think I complicated it by putting in the values of P1 and P0 without simplifying it first. Also, I didn't think of simplifying by using 2x10³. I ended up with huge numbers which confused me more.