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The currrent, I0 (measured in amperes), decreases to I amperes after t seconds according to the formula I = I0e−0.51t. When 8 amperes decreases to 3.2 amperes, the equation is 3.2 = 8e−0.51t. Rounded to the nearest tenth of a second, how long, t, does it take for the current to decrease from 8 amps to 3.2 amps?

A. t = 0.4 second

B.t = 1.8 seconds

C. t = 2.5 seconds

D.t = 0.7 second

 Oct 21, 2017

Best Answer 

 #1
avatar+27529 
+1

3.2 = 8e-0.51t  

 

Take logs of both sides

 

ln(3.2) = ln(8e-0.51t)

 

Use the properties of logs to express this as

 

ln(3.2) = ln(8) - 0.51t

 

Rearrange:

 

0.51t = ln(8) - ln(3.2)

 

0.51t = ln(8/3.2)

 

t = ln(8/3.2)/0.51 ≈ 1.8 s

 Oct 21, 2017
 #1
avatar+27529 
+1
Best Answer

3.2 = 8e-0.51t  

 

Take logs of both sides

 

ln(3.2) = ln(8e-0.51t)

 

Use the properties of logs to express this as

 

ln(3.2) = ln(8) - 0.51t

 

Rearrange:

 

0.51t = ln(8) - ln(3.2)

 

0.51t = ln(8/3.2)

 

t = ln(8/3.2)/0.51 ≈ 1.8 s

Alan Oct 21, 2017

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