My Math Forum Finding the value of A and b from a log graph.

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 December 3rd, 2017, 02:31 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 2-3 questions. I believe "A'' is 1,9 because for the differential equation I took: 0,74-0,175 = 0,565 and 0,003-0,298 = 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 02:57 PM.
 December 3rd, 2017, 03:10 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,870 Thanks: 1833 Is 0,45 in the table supposed to be 0,55? Were you given the table or the graph? Your equation "0,003-0.298 = 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 03:38 PM.
 December 3rd, 2017, 03:24 PM #3 Newbie   Joined: Dec 2017 From: USA Posts: 3 Thanks: 0 Yes, 0,55 my apologies. I was given the non-log 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 04:03 PM.
 December 3rd, 2017, 04:05 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,870 Thanks: 1833 There are minor inaccuracies in your log table. Were you told to use logs?
December 3rd, 2017, 04:18 PM   #5
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Quote:
 Originally Posted by skipjack There are minor inaccuracies in your log table. Were you told to use logs?
True; I rounded it up a bit, but it shouldn't matter much. I just need to learn the steps on how to find A and b.

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 01:43 PM.

 December 4th, 2017, 03:06 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,870 Thanks: 1833 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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