Logs in Equations

Date: 05/20/97 at 12:38:38
From: Hughie Coyle
Subject: Calculation of Output Power 

I have already calculated that a circuit has an attenuation of 60dB's
and the input power = 200mW and I need to calculate the mean output 
power.

The formula is as follows :

		 dB = 10log PowerOut/PowerIn

		-60 = 10log PowerOut/200/1000


The question is: What do I have to do to manipulate this formula in 
order to find the PowerOut?

In order to get PowerOut on its own, I tried taking the natural log
of both sides, but the answer I get is incorrect (818mW).

Can you please enlighten me as to the correct way of manipulation of
logs?

Best Regards,
H F Coyle

Date: 06/24/97 at 14:49:00
From: Doctor Barney
Subject: Re: Calculation of Output Power Rev. 1

Hi Hughie,  

I would be happy to spread some enlightenment.  Let's review what a 
logarithm is.  

If you have a number B, and you raise it to the power of x, and the 
result is C, then B^x = C. We say that the log (base B) of C is x:  

log(C) = x   

Now, how can you get rid of the log in an equation?  Suppose we take 
our base number, B, and raise it to the power of both sides of that 
last equation:

B^(log(C)) = B^x.

But B^x = C, so  B^(log(C)) = C.

This is sometimes called an inverse logarithm.  In plain English, to 
get rid of the log, raise it to the power of whatever base you are 
using, and the answer is the thing you were taking the log of in the 
first place.   

You can have logarithms of any base you want, but two of the most 
common are logs with a base of e, or natural logarithms, and 
logarithms with a base of 10, or common logarithms. A decibel is an 
engineering unit that was created to make it more convenient to 
describe ratios of very different power levels. In the formula you 
site above, which is the definition of a decibel, the log is a base 10 
logarithm. (This problem has nothing to do with natural logarithms or 
with the number e.) As an example of how we might use a decibel, if 
the ratio of the power of the signal coming out of your stereo 
amplifier to the power going in is a thousand to one, that's 30 dB:

     10 x log(1,000/1) = 30   

In the problem you have above, first write the problem like this:

-60 = 10log(PowerOut/PowerIn)

-60 = 10 x log(PowerOut/200)

If you leave the input power in mW, the output power will also be in 
mW. First isolate the log by dividing both sides of the equation by 
10. Then, to get rid of the log, raise 10 to the power of both sides 
of the equation. 
 
Then it should be easy to solve for the output power. Since this 
circuit has an attenuation of 60 dB, your answer should be much, much 
smaller than 200 mW.   

-Doctor Barney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/