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November 27th, 2014, 12:53 PM   #1
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How did they get this answer?

L=10log(80/1x10^-12)

L=139.03dB
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November 27th, 2014, 01:32 PM   #2
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What is "dB"?
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November 27th, 2014, 01:51 PM   #3
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What is "dB"?
Decibel.

Last edited by greg1313; November 27th, 2014 at 01:54 PM.
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November 27th, 2014, 01:58 PM   #4
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Evaluating with a calculator, $\displaystyle 10\log\left(\frac{80}{10^{-12}}\right)\approx139.03$
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Last edited by greg1313; November 27th, 2014 at 02:17 PM.
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November 27th, 2014, 01:59 PM   #5
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Evaluating with a calculator, $\displaystyle 10\log\left(\frac{80}{10^{-12}}\right)\approx139.08$
Cheers bud, I am an idiot.
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November 27th, 2014, 02:01 PM   #6
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Quote:
Originally Posted by SeanDaniel View Post
L=10log(80/1x10^-12)

L=139.03dB
$\displaystyle L=10\log( 80)-10\log(10^{-12})$

$\displaystyle L=10\log( 8 )+10\log(10)-10\log(10^{-12})$

$\displaystyle L=10\log( 8 )+10+120$

$\displaystyle L=9.03+130$ decibels
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November 27th, 2014, 02:08 PM   #7
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Quote:
Originally Posted by SeanDaniel View Post
L=10log(80/1x10^-12)

L=139.03dB
SeanDean, you need grouping symbols around the denominator.

Here, "log" is taken to be to base 10.


"dB" stands for "decibels."


L = 10log{80/[1 x 10^(-12)]} dB

L = 10log{80/[10^(-12)]} dB

L = 10log(80*10^12) dB

L = 10log(8*10*10^12) dB

L = 10log(8*10^13) dB

L = {10[log(8) + log(10^13)]} dB

L = {10log(8) + 10log(10^13)} dB

L = [10log(8) + 10(13)] dB

L ~ [10(0.90309) + 130] dB

L ~ (9.0309 + 130) dB

L = 139.03 dB $\displaystyle \ \ \ $when rounded to two decimal places


Quote:
Originally Posted by greg1313 View Post
Evaluating with a calculator, $\displaystyle 10\log\left(\frac{80}{10^{-12}}\right)\approx139.08$
That calculator is off, or possibly you copied that last digit incorrectly.
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Last edited by Math Message Board tutor; November 27th, 2014 at 02:11 PM.
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November 27th, 2014, 02:17 PM   #8
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Originally Posted by Math Message Board tutor View Post
That calculator is off, or possibly you copied that last digit incorrectly.
I mis-keyed '80' as '81', thanks.
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