Calculus help?


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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...

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** 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}

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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

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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

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.

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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...

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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.

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 by SanGreal
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