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November 11th, 2007, 12:31 PM   #1
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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*(x-2)) = e
2: e^x > e^-x
3: (1/5)^(x²) < 5^(-12-4x)
4: ln(x)-ln(x-4) = 4
5: ln(1-x) < 0

how far i could get by myself:
1: e^(x² -2x) = e
x²-2x = 1
x² -2x -1 = 0
abc formula:
2+-(-2²-4*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^(-12-4x)
5^(-x²) < 5^(-12-4x)
-x² < -12-4x
0 < x²-4x-12
(x-6) (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(x-4) = 4
ln(x/(x-4)) = 4
x/(x-4) = e^4
x=e^4*(x-4)
x=e^4x-4e^4
4e^4=e^4x+x
and thats how far i could get

5: ln(1-x) < 0
1-x < 1
0 < x
but according to the book the answer is 1>x>0
help..
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November 11th, 2007, 01:07 PM   #2
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Hello,

1) well, (-2)²-4*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(1-x)<0
then, below, 1-x<1
in fact you didn't wrote an equivalence but only an implication, because ln(1-x) doesn't make any sense for x>=1 (so neither ln(1-x)<0), while 1-x does (so 1-x<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(1-x)<0, you should have wrote 1-x<1 and x<1


PS : sorry for my pitiful english. Do not hesitate to correct my mistakes.
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November 11th, 2007, 01:12 PM   #3
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(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/a-b^2/(2a)^2=0
(x+b/(2a))^2=b^2/(4a^2)-c/a
(x+b/(2a))^2=(b^2-4ac)/(4a^2)
x+b/(2a)=±(√(b^2-4ac))/2a
x=(-b±√(b^2-4ac))/(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^(-12-40)

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(1-x)<0 you need to have:

1-x<1
or
x>0. But, also:
1-x>0 (the conclusion about ln(x) being real)
1>x

this would yield your needed soluton 1>x>0
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November 11th, 2007, 01:18 PM   #4
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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...").
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November 11th, 2007, 01:34 PM   #5
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Quote:
Originally Posted by Le barbu rasé
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...").
oh ok, im sorry. merci beaucoup by the way
ur english is fine, you know im dutch and my girlfriend is french
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November 11th, 2007, 02:04 PM   #6
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Croatia here.. ironic..
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November 20th, 2007, 11:41 AM   #7
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November 20th, 2007, 11:48 AM   #8
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Quote:
Originally Posted by Le barbu rasé
PS : sorry for my pitiful english. Do not hesitate to correct my mistakes.
I wish most of our our native English speakers wrote as well as you. I wouldn't sweat it, your English is good. Far better than my weak French, I assure you. Je lis le francais mais up peut, et je ne se parle jamais.
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November 21st, 2007, 12:32 AM   #9
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Haha.. It's either your typing or your French! And I don't mean this in a bad way, don't get insulted CRGreathouse!
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November 21st, 2007, 03:50 AM   #10
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Quote:
Originally Posted by milin
Haha.. It's either your typing or your French! And I don't mean this in a bad way, don't get insulted CRGreathouse!
Actually I was relieved -- at first I thought you were saying that my French was incomprehensible (to a French speaker) instead of just having bad grammar... but now I see you mean you just don't speak the language. Much better.

Of course I did leave out my accents, so the text does look ugly. Ah, c'est la vie...
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