# SOLUTION: The attenuation of a 900 MHz mobile phone signal by a concrete wall reinforced with steel mesh is described by, I(t)=1.2e^(-0.138t) where t is the thickness of the wall in c

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: The attenuation of a 900 MHz mobile phone signal by a concrete wall reinforced with steel mesh is described by, I(t)=1.2e^(-0.138t) where t is the thickness of the wall in c      Log On

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 Click here to see ALL problems on Exponential-and-logarithmic-functions Question 607998: The attenuation of a 900 MHz mobile phone signal by a concrete wall reinforced with steel mesh is described by, I(t)=1.2e^(-0.138t) where t is the thickness of the wall in cm, and I(t) is the intensity of signal, in mW/m2, measured on the interior of the wall. (a) What is the signal intensity incident on the exterior of the wall? (a) Answer - I think is t=0 900=1.2e^1.2^-0.138*0 900=1.2*1 900/1.2=1.2*1/1.2 750=1 750/1=0 750 (b) If the wall is 25 cm thick, what intensity will be measured inside? Round your answer to 2 decimal places. (b) Answer - I think is 750=1.2e^-0.138*0.25 750=1.2e^-0.0345 750/1.2*0.966088=1.2*0.966088/1.2*0.966088 646.9389952 Approx =647 (c) What thickness of wall will drop the intensity to half of the exterior value? Round your answer to 2 decimal places. (c) Not sure a bit lost Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!you need to watch what values you substitute where... (a) I(0) = 1.2 e^[-0.138(0)] = 1.2 e^0 = 1.2 (b) I(25) = 1.2 e^[-0.138(25)] (c) 1/2 = e^[-0.138(x)] ___ ln(1/2) = -0.138 x