Converting sound intensity to dB

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The sound intensity at a concert is measured at 0.5 W/m^2, leading to confusion over the correct decibel calculation. The initial attempt used the equation β = 10ln(I/I_0), resulting in an incorrect high value of 269.378 dB. The correct formula for sound intensity in decibels is β = 10*log10(I/I_0), which yielded a more expected result of 116.99 dB. The user clarified that they initially confused logarithmic functions, realizing that log(x) is not the same as ln(x). The discussion also included a query about simplifying the log function on the TI-89 calculator.
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Homework Statement



The sound at a concert is measured to be 0.5 W/m^2. How many decibels is this?

Homework Equations



β = 10ln(I/I_0)
I/I_0 = 10^(β/10)

The Attempt at a Solution



I_0 = 1*10^-12 W/m^2
I = 0.5 W/m^2

Therefore,
β = 10ln(0.5/10^-12)
β = 269.378 dB

However,
.5/10^12 = 10^(β/10)
When using a TI-89 calculator to solve for β the answer comes to 116.99 dB, what I expected.

I can't find anything wrong with my work, but the dB level seems way too high for the first equation. I expected it should be somewhere around 115 dB. Why does the formula involving natural logs give the wrong answer?
 
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The sound intensity in dB is defined as 10*log10(I/I0).

ehild
 
ehild said:
The sound intensity in dB is defined as 10*log10(I/I0).

ehild

Nevermind, I thought log(x) was another way to write ln(x)... some research cleared this up. However, is there an easier way to find the log( function in the TI-89 calculator other than scrolling through the catalog?
 
Last edited:
log(x)=ln(x)/ln(10). Calculate ln(I/I0) and divide by ln(10)=2.302.

ehild
 
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