help with logarithms
| Author | Topic: help with logarithms |
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Warrior
Member # 5483
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written Monday, December 18 2006 12:31
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I missed a day and I don't know how to solve these types of problems (I have a test on them tomorrow). While these are homework, it isn't graded and the answers are in the back of the book so don't think I'm cheating. I just need to know how to do them. In {} are subscripts. Evaluate. 29. 2log{3}6 - log{3}4 31. log{4}40 - log{4}5 Here is part of a different set: Solve each equation. 35. log{a}x = 3/2log{a}9 + log{a}2 39. log{a}(3x+5) - log{a}(x-5) = log{a}8 41. log{3}(x+2) + log{3}6 = 3 no subscripts below this line answers: 29. 2 31. 3/2 35. {54} 39. {9} 41. 5/2 thanks A LOT in advance :D -------------------- Ignorance Is bliss -Cypher (Matrix) Don't think you can; know you can -Morpheus (Matrix) sanity is overrated :) Posts: 130 | Registered: Monday, February 7 2005 08:00 |
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Law Bringer
Member # 335
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written Monday, December 18 2006 12:41
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The first thing you need to be able to do is convert from logs with strange bases (let's call them a) to familiar bases (let's call them b, but you'll usually use either e or 10). Fortunately, there's a formula: Log{n}(x) = log{b}(x) / log{b}(n) So log{3}6 = ln(6)/ln(3) [Edit: Whoops, didn't finish.] Other things to remember are that a*ln(b) = ln(b^a) and ln(a) + ln(b) = ln(a*b). You can use these to simplify equations into one log{n}(something) = log{n}(something else} and then cancel logs. Again, an example: log(x) = 3/2log(9) + log(2) log(x) = log[ 9^(3/2) * 2] x = 2*9^(3/2) x = 54 Finally , you can always turn plain old numbers into logarithms so you can combine terms. k = log{a}(k^a) Ex: 5 = log{2}(25) —Alorael, who is logging out in the manner of Thuryls everywhere. [ Monday, December 18, 2006 13:10: Message edited by: Elegance:: Qwghlmian or Lojban? ] Posts: 14579 | Registered: Saturday, December 1 2001 08:00 |
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Post Navel Trauma ^_^
Member # 67
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written Monday, December 18 2006 12:42
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log{a}x - log{a}y = log{a}(x/y) y log{a}x = log{a}(x^y) if x = a^y then y = log{a} x You don't need to convert into other bases, for these questions. [ Monday, December 18, 2006 12:43: Message edited by: Khoth ] -------------------- Barcoorah: I even did it to a big dorset ram. desperance.net - Don't follow this link Posts: 1798 | Registered: Thursday, October 4 2001 07:00 |
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Shock Trooper
Member # 6666
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written Monday, December 18 2006 12:55
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Also: log(x)+log(y) = log (x*y) EDIT: dammit, Alo did this one already. I'm useless. Ah well, here are the answers to the questions above. No peaking before you've tried them yourself, and so forth. One thing about my answers: I use the line | to separate the formula from different rules and other types of manipulation I apply to the formula. It's pretty common in Finland, but from what I've understood, not necessarily used abroad. 29. 2log{3}6 - log{3}4 | x log y = log(y^x) log{3}36 - log{3}4 | log x - log y = log (x/y) log{3}9 | log{x}y = z <=> x^z = y 3^x = 9 3^x = 3^2 x = 2 31. log{4}40 - log{4}5 log{4}8 4^x = 8 log{2}(4^x) = log{2}8 | log (x^y) = y log x x log{2}4 = log{2}8 x * 2 = 3 x = 1,5 35. log{a}x = 3/2log{a}9 + log{a}2 log{a}x = log{a}(9^(3/2) * 2) x = 9^(3/2) * 2 x = 54 39. log{a}(3x+5) - log{a}(x-5) = log{a}8 log{a}((3x+5)/(x-5)) = log{a}8 3x+5 = 8x-40 45 = 5x x = 9 41. log{3}(x+2) + log{3}6 = 3 log{3}(x+2) = log{3}27/6 x+2 = 27/6 x = 27/6-2 log{3}(6x+12) = 3 | 3 = log{3}27 log{3}(6x+12) = log{3}27 6x + 12 = 27 6x = 15 x = 5/2 [ Monday, December 18, 2006 13:08: Message edited by: Redstart ] Posts: 353 | Registered: Monday, January 9 2006 08:00 |