Longsiege Posted January 11, 2004 Share Posted January 11, 2004 Ok I have a calculus paper which contains 4 questions. Anyways I solved the first three with minimal difficulty however the last question I don't even know how to start gonig about it. If anyone can solve it or point me to somewhere or someone who can I'd be extremely greatfull... The Question... Determine the real values of x, for which F is positive where F = log(a) (2x+4/3x) a > 0, a doesn't equal 1... Link to comment Share on other sites More sharing options...
lardiop Veteran Posted January 11, 2004 Veteran Share Posted January 11, 2004 Google a place called "Dr. Math"... like a forum for math questions. Link to comment Share on other sites More sharing options...
Longsiege Posted January 11, 2004 Author Share Posted January 11, 2004 dr math is terrible lol... Link to comment Share on other sites More sharing options...
m0fo Posted January 11, 2004 Share Posted January 11, 2004 ** i'm assuming you typed the q correctly and it is 2x + (4/3x) and not (2x + 4) / 3x ** to prove that F is positive, you have to prove that (2x+4/3x) is greater than 1 so, (2x + 4/3x) > 1 ** note that x != 0 - restriction** multiply every term by 3x: 6x^2 + 4 > 3x 6x^2 - 3x + 4 > 0 x = (3 ? sqrt(9 - 96)) / 12 x = (3 ? sqrt(87)i) / 12 therefore, x < {[3 - sqrt(87)i] / 12} or x > {[3 + sqrt(87)i] / 12} Link to comment Share on other sites More sharing options...
SanGreal Posted January 11, 2004 Share Posted January 11, 2004 to prove that F is positive, you have to prove that (2x+4/3x) is greater than 1 Why? if (2x+4/3x) were 1, wouldn't F still be positive? log(a) will always be positive Link to comment Share on other sites More sharing options...
m0fo Posted January 11, 2004 Share Posted January 11, 2004 if (2x+4/3x) = 1, then F would be equal to 0.... a^0 = 1 log(a) is only positive when (2x+4/3x) is greater than 1... say that a=2 and (2x+4/3x) = 1/2, then log(2)(1/2) = -1 and therefore not positive Link to comment Share on other sites More sharing options...
Longsiege Posted January 11, 2004 Author Share Posted January 11, 2004 ** i'm assuming you typed the q correctly and it is 2x + (4/3x) and not (2x + 4) / 3x ** It is typed correctly :p i'll scan it for you... Link to comment Share on other sites More sharing options...
m0fo Posted January 11, 2004 Share Posted January 11, 2004 no, u didnt' type it correctly... it should be [(2x + 4)/3x] not (2x + 4/3x) in this case, solve for (2x + 4) / 3x > 1 and so, 2x + 4 > 3x 4 > x x < 4 (x != 0) Link to comment Share on other sites More sharing options...
Longsiege Posted January 11, 2004 Author Share Posted January 11, 2004 Thanx for the help though mofo, i'm not sure if it solves it, but its a start :p.. Link to comment Share on other sites More sharing options...
Longsiege Posted January 11, 2004 Author Share Posted January 11, 2004 Ah ok I get it, Thanx you very much! Great help! Link to comment Share on other sites More sharing options...
SanGreal Posted January 11, 2004 Share Posted January 11, 2004 if (2x+4/3x) = 1, then F would be equal to 0.... a^0 = 1log(a) is only positive when (2x+4/3x) is greater than 1... say that a=2 and (2x+4/3x) = 1/2, then log(2)(1/2) = -1 and therefore not positive er, log(2)(1/2) is not -1 As long as a is positive, as defined in the question, log(a) should come out positive. Link to comment Share on other sites More sharing options...
Longsiege Posted January 11, 2004 Author Share Posted January 11, 2004 uh so his answer is wrong?... I can prove X < 4.. but I think the domian is 0 < x < 4 for some reason... I don't know how to prove the x > 0 though... Link to comment Share on other sites More sharing options...
SanGreal Posted January 11, 2004 Share Posted January 11, 2004 (edited) er, log(2)(1/2) is not -1As long as a is positive, as defined in the question, log(a) should come out positive. nm; I didn't realize it was log base a of [(2x+4)/3x]. I thought it was log base 10 of a * [(2x+4)/3x] he is definately right (and so are you), the answer is 4 > x > 0 x can't be 0 because 3(0) is 0 which makes (2x+4)/3x undefined. Even if you split it up as log(a)2x+4 - log(a)3x it still doesn't work because log(a)0 is undefined. oh, and you can't take a log of a negative number, thats why you can't have less than 0 Edited January 11, 2004 by SanGreal Link to comment Share on other sites More sharing options...
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