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December 3rd, 2017, 01:31 PM  #1 
Newbie Joined: Dec 2017 From: USA Posts: 3 Thanks: 0  Finding the value of A and b from a log graph.
This is the question in my book: https://image.prntscr.com/image/eoWE...tBk7l6Z6WQ.png I made on paint how far I got so far. However I'm having trouble with the last 23 questions. I believe "A'' is 1,9 because for the differential equation I took: 0,740,175 = 0,565 and 0,0030,298 = 0,298 and 0,565/0,298 = 1,9 However, can someone verify that that's correct? And if someone could help me with the other two, I'll very much appreciate it. Thank you. Last edited by skipjack; December 3rd, 2017 at 01:57 PM. 
December 3rd, 2017, 02:10 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,299 Thanks: 1688 
Is 0,45 in the table supposed to be 0,55? Were you given the table or the graph? Your equation "0,0030.298 = 0.298" should be "0,003  (0.295) = 0.298". Try to be consistent with the variable names... is it U or V, a or A, b or B? I would think that A = 1,9 is close enough. The value of B should be slightly less than 0,74 (anything from 0,734 to 0,737 would be acceptable). Last edited by skipjack; December 3rd, 2017 at 02:38 PM. 
December 3rd, 2017, 02:24 PM  #3 
Newbie Joined: Dec 2017 From: USA Posts: 3 Thanks: 0 
Yes, 0,55 my apologies. I was given the nonlog table V, 1.5 2,5 3,5 4,5 5,5 I, 0,51 0,66 0,79 0,9 1,005 Last edited by skipjack; December 3rd, 2017 at 03:03 PM. 
December 3rd, 2017, 03:05 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,299 Thanks: 1688 
There are minor inaccuracies in your log table. Were you told to use logs?

December 3rd, 2017, 03:18 PM  #5  
Newbie Joined: Dec 2017 From: USA Posts: 3 Thanks: 0  Quote:
Yes. We got a standard table. Turn it into log. Draw a graph, put the dots on it, and then solve for A and B to make the formula U = b*I^a, so all I need is A and b. Last edited by skipjack; December 4th, 2017 at 12:43 PM.  
December 4th, 2017, 02:06 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,299 Thanks: 1688 
Having found a = 1.9 correctly, b = 0.74  1.9*0.003 = 0.7343. Using b = 0.175 + 1.9*0.295 would give about the same result, especially if you hadn't rounded up the logarithms. 

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finding, graph, log 
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