Expert: Sherman D. - 3/19/2004

Question
The question we received in class is as followed along with the  work i've used to solve it. Will you please inform me if I'm doing anything wrong?

Consider the equation Pr= Po(d/do)^-n.  Determine the value of n if the received power at 13.8km is Pr= -10dBm, if the reference distance is 123 meters and the reference power measured at that distance is 10dBm.

My work:

Pr/Po = (d/do)^-n
-1 = (13.8/123)^-n
-1 = (0.0112195) ^-n
1= -(0.0112195)^-n

ln1 = +n ln (0.0112195)

n=0

PLEASE HELP!! Thanks again!

Answer
Pr = Po(d/do)^-n

divide both sides by Po

Pr/Po = (d/do)^-n

same as

Pr/Po = 1/((d/do)^n)

Multiply both sides by (d/do)^n

((d/do)^n)(Pr/Po) = 1

divide both sides by (Pr/Po)

(d/do)^n = 1/(Pr/Po)

using this

b^y = x means log(b)x = y

You get

n = log(base(d/d0))(1/(Pr/Po))

Now according to your substitutions, you get

n = log(base(13.8/123))(-1)

which becomes

n = log(base(138/1230))(-1)
n = log(base(23/205))(-1)

That is what i get.

and i believe "ln" is only involved if "e" is involved.

I could be wrong about "ln" and "e" since it has been while since i did logarithmic and exponential problems.

I am just taking that from the sites
http://www.chem.tamu.edu/class/fyp/mathrev/mr-log.html
and
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/logs.html