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November 27th, 2014, 01:53 PM  #1 
Newbie Joined: Nov 2014 From: London Posts: 21 Thanks: 0  How did they get this answer?
L=10log(80/1x10^12) L=139.03dB 
November 27th, 2014, 02:32 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
What is "dB"?

November 27th, 2014, 02:51 PM  #3 
Newbie Joined: Nov 2014 From: London Posts: 21 Thanks: 0  Last edited by greg1313; November 27th, 2014 at 02:54 PM. 
November 27th, 2014, 02:58 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
Evaluating with a calculator, $\displaystyle 10\log\left(\frac{80}{10^{12}}\right)\approx139.03$
Last edited by greg1313; November 27th, 2014 at 03:17 PM. 
November 27th, 2014, 02:59 PM  #5 
Newbie Joined: Nov 2014 From: London Posts: 21 Thanks: 0  
November 27th, 2014, 03:01 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,774 Thanks: 1428  
November 27th, 2014, 03:08 PM  #7 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  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 That calculator is off, or possibly you copied that last digit incorrectly. Last edited by Math Message Board tutor; November 27th, 2014 at 03:11 PM. 
November 27th, 2014, 03:17 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond  

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