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November 11th, 2007, 12:31 PM  #1 
Newbie Joined: Nov 2007 Posts: 2 Thanks: 0  need help, logarithms
I've got a problem with some math exercises * I cant use a calculator, i must answer like this for example: x = ln(3) what is x? 1: e^(x*(x2)) = e 2: e^x > e^x 3: (1/5)^(x²) < 5^(124x) 4: ln(x)ln(x4) = 4 5: ln(1x) < 0 how far i could get by myself: 1: e^(x² 2x) = e x²2x = 1 x² 2x 1 = 0 abc formula: 2+(2²4*1*1)^0.5 2 x=2,41 v x=0,41 I cant do this out of my head, according to the book the answer was 1+ (2)^0.5, how did they do it? 2:e^x > e^x I thought this was simply x > x , but it turned out to be x > 0 why is it x > 0? and was my answer correct also ? 3: (1/5)^(x²) < 5^(124x) 5^(x²) < 5^(124x) x² < 124x 0 < x²4x12 (x6) (x+2) x=6 v x=2 ok im not sure if its x<6 or x>6, or x<2 or x>2. Im always doing this wrong, could somebody explain how it works? 4: ln(x)ln(x4) = 4 ln(x/(x4)) = 4 x/(x4) = e^4 x=e^4*(x4) x=e^4x4e^4 4e^4=e^4x+x and thats how far i could get 5: ln(1x) < 0 1x < 1 0 < x but according to the book the answer is 1>x>0 help.. 
November 11th, 2007, 01:07 PM  #2 
Newbie Joined: Sep 2007 From: Nice, France Posts: 13 Thanks: 0 
Hello, 1) well, (2)²4*1*(1)=8 and √8=2*√2. Then just simplify. 2) x>x is correct, but it doesn't gives us any idea of where x belongs to. But x>x <=> x+x>0, right ? 3) The rule is that a second degree polynomial has the sign of its leading coefficient outside of the roots. 4) well, you have now an equation of the form ax=b. easy. 5) well, when you write ln(1x)<0 then, below, 1x<1 in fact you didn't wrote an equivalence but only an implication, because ln(1x) doesn't make any sense for x>=1 (so neither ln(1x)<0), while 1x does (so 1x<1 too, and worse, it even can be true, and worst, a posteriori, here, it is!). When you want to solve an inequation, you can't really avoid to proceed by equivalences. So, below ln(1x)<0, you should have wrote 1x<1 and x<1 PS : sorry for my pitiful english. Do not hesitate to correct my mistakes. 
November 11th, 2007, 01:12 PM  #3 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
(1) To get the quadratic equation roots, try this: a x^2 + b x + c = 0 x^2 + b/a x + c/a = 0 (x+b/(2a))^2+c/ab^2/(2a)^2=0 (x+b/(2a))^2=b^2/(4a^2)c/a (x+b/(2a))^2=(b^24ac)/(4a^2) x+b/(2a)=±(√(b^24ac))/2a x=(b±√(b^24ac))/(2a) so, you have: a=1 b=2 c=1 and you get: x=1±√2 (2) well, x>x is x+x>0 and that's 2x>0 > x>0 (3) you could always try! there are only 3 possible solutions: x E (∞,2) x E (2,6) x E (6,+∞) so, for x=3 you get: 5^(9)<5^(12+12) 5^(9)<1 and that's true! now, for x=0, you get: 5^0<5^(12) this is not true. And finally, for x=10 you get: 5^(100)<5^(1240) and this is true as well. So, your solution would be: x E (∞,2)U(6,+∞) (4) 4e^4=e^4x+x 4e^4=(e^4+1)x 4e^4/(e^4+1)=x (5) you know that ln(x) is real only for x>0. Also, you know that ln(1) is 0, ln(x)>0, x>1 and ln(x)<0, x<1 so, for ln(1x)<0 you need to have: 1x<1 or x>0. But, also: 1x>0 (the conclusion about ln(x) being real) 1>x this would yield your needed soluton 1>x>0 
November 11th, 2007, 01:18 PM  #4 
Newbie Joined: Sep 2007 From: Nice, France Posts: 13 Thanks: 0 
By the way, I'm not sure it was the right section to post your answer (I would have say something like "Real Analysis..." or "Calculus...").

November 11th, 2007, 01:34 PM  #5  
Newbie Joined: Nov 2007 Posts: 2 Thanks: 0  Quote:
ur english is fine, you know im dutch and my girlfriend is french  
November 11th, 2007, 02:04 PM  #6 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
Croatia here.. ironic..

November 20th, 2007, 11:41 AM  #7 
Newbie Joined: Sep 2007 From: Nice, France Posts: 13 Thanks: 0  
November 20th, 2007, 11:48 AM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 
November 21st, 2007, 12:32 AM  #9 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
Haha.. It's either your typing or your French! And I don't mean this in a bad way, don't get insulted CRGreathouse! 
November 21st, 2007, 03:50 AM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Of course I did leave out my accents, so the text does look ugly. Ah, c'est la vie...  

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