Logarithms  Show that the expression is trueby JJBladester Tags: decibel, expression, logarithm, true 

#1
Dec1412, 04:26 PM

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P: 280

1. The problem statement, all variables and given/known data
Show that the following expression is true: 0.2*0.8 = 14dB + (1.94dB) 2. Relevant equations 1 dB = 10log_{10}(x) (where x is a ratio of two quantities) 10log_{10}(ab) = 10log_{10}(a) + 10log_{10}(b) 3. The attempt at a solution 0.2*0.8 = 14dB + (1.94dB) 0.16 = 15.94dB ___________________________ 10log_{10}(x)=15.94 x=10^{15.94/10}=0.0255 ___________________________ 0.16 ≠ 0.0255 (so either my math is wrong or the expression is not true...) 



#2
Dec1412, 04:45 PM

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P: 20,988

Is this actually how the problem is presented? 



#3
Dec1412, 04:49 PM

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P: 280





#4
Dec1412, 05:00 PM

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P: 280

Logarithms  Show that the expression is true
On your note about units... Decibels are unitless (dimensionless)... So aside from that, am I missing something else?




#5
Dec1412, 05:12 PM

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P: 20,988

I guess that the idea is to convert each dB value using the conversion formula, and see if the two dB values add up to .16.




#6
Dec1412, 09:16 PM

P: 1,784

JJBladestr, let's start with the logarithm of 0.2. What do you get for that?
What do you get for the logarithm of 0.8? Now let's multiply each of those by 10, not 20. Add the dB values together. and take the antilog of the sum. 



#7
Dec1512, 08:04 AM

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P: 280

10log_{10}(0.8) = 0.9691 dB 6.9897 dB + 0.9691 dB = 7.9588 dB 10^{7.9588/10}=0.16 But that has not proven anything about the left side equaling the right side. It has merely manipulated the left side and we are back to the original question. 



#8
Dec1512, 12:36 PM

P: 1,784

You're right. Where did the 14 dB and the 1.94 dB come from? If they are part of the problem then either they are not equal or you should have used 20*Log instead of 10*Log, but I see nothing in the problem to indicate that.




#9
Dec1512, 04:06 PM

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P: 280

I asked the professor what was going on with the problem and he said that the math I did was right but the problem was not formed properly.
The thing with online schools is some of them have a group of staff members who create the homework/labs/tests and a separate group who instructs. This is an ABETaccredited school, which is good because I'm going to be an Electrical Engineer at the end of it all. I see what you mean about 20*log(something). The "20" would indicate that two voltage levels were being compared because the formula for voltage gain has a "20" in it. 


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