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Natural logarithmic equation help s'il vous plait Watch

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    "Prove, by counter-example, that the statement
    “ln (3x 2 + 5x + 3) ≥ 0 for all real values of x” is false."

    Hmmm I'm not quite sure how to go about this so i'll post my working below & could you check it please?

    Also can some explain what a real number is? I don't do further maths but I'm assuming it's just a whole integer.. So for example, -2/3 isn't a real number...?

    Thanks
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    A real number is just what you would normally call a number. Anything on the number line, so yes, integers, but also fractions, surds, etc. As you suspect, it is only in further maths that you meet numbers that are not on the usual number line.

    Note that you are not asked to find a value of x that is <0. You are asked to find a value of x for which ln(3x^2+5x+3)<0. When you put in either of the x values you have found, the expression is equal to 0, not less than it. However - can you think of a way to use the two values you have found to suggest another value of x that will fit the bill?
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    A real number is one that does exist (root -1 isn't a real number).

    Its asking for an example that disproves this statement. So to disprove it you need to show the quadratic being equal to a value that makes the value of the natural log being smaller than zero and show that there are real values of x (basically the quadratic equation works). Natural log is negative when the function inside is between 0 and 1.
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    Actually, on reflection, I can't see an approach based on your idea that will get you to the end, so forget the bit I wrote about trying to use those answers.

    I would suggest completing the square on the expression you start off with and seeing if that gives you any ideas.
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    (Original post by Pangol)
    Actually, on reflection, I can't see an approach based on your idea that will get you to the end, so forget the bit I wrote about trying to use those answers.

    I would suggest completing the square on the expression you start off with and seeing if that gives you any ideas.
    To be completely honest so far to work this out I spammed decimals into my calculator until it worked with an example.
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    (Original post by Vikingninja)
    To be completely honest so far to work this out I spammed decimals into my calculator until it worked with an example.
    I guess that would be OK, but if you complete the square, there is one particular value for x that is crying out to be the ideal counterexample.
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    (Original post by jessyjellytot14)
    "Prove, by counter-example, that the statement
    “ln (3x 2 + 5x + 3) ≥ 0 for all real values of x” is false."

    Hmmm I'm not quite sure how to go about this so i'll post my working below & could you check it please?

    Also can some explain what a real number is? I don't do further maths but I'm assuming it's just a whole integer.. So for example, -2/3 isn't a real number...?

    Thanks
    That's not a proof, it doesn't disprove the conjecture whatsoever. You simply found the roots of the equation equalling 0 whereas you want to find a value of x for which \ln(3x^2+5x+3)&lt;0 as then the statement would be false if such real x value exists (a contradiction arises concerning the domain stated in the conjecture)

    To properly disprove it, think about \ln(a)\geq 0 and how a must be greater than or equal 1 for this to be true. So you want to show that \ln(a)&lt;0 exists therefore you want to solve a&lt;1 \Rightarrow 3x^2+5x+3&lt;1

 and show that there exists x values for which 3x^2+5x+3 is less than 1 - hence there exist x values for which \ln(3x^2+5x+3)&lt; 0

    P.S. If you have no knowledge of complex numbers then a real number is just any number you can think of.
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    (Original post by Vikingninja)
    A real number is one that doesn't exist (root -1 isn't a real number).
    (I know that you mean "does exist" when you say "doesn't exist" - what I'm about to say isn't supposed to be a comment on the kind of typo that we all make.)

    I would claim that complex numbers do exist! The square root of -1 "exists" in the same way that 2, 7/11 and pi "exist" - it's just not real. But this is more of a philosophical discussion that is getting OT...
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    (Original post by Pangol)
    (I know that you mean "does exist" when you say "doesn't exist" - what I'm about to say isn't supposed to be a comment on the kind of typo that we all make.)

    I would claim that complex numbers do exist! The square root of -1 "exists" in the same way that 2, 7/11 and pi "exist" - it's just not real. But this is more of a philosophical discussion that is getting OT...
    Yeah I just kept it as that because of all the weird stuff around (LOVE it at uni, really fun especially in engineering).
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    (Original post by Pangol)
    I guess that would be OK, but if you complete the square, there is one particular value for x that is crying out to be the ideal counterexample.
    Okay I've tried to complete the square but it doesn't seem to work...
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    Not sure if this is what you meant
    I didn't finish it because I knew I'd just get a math error if I rooted it.

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    (Original post by RDKGames)
    That's not a proof, it doesn't disprove the conjecture whatsoever. You simply found the roots of the equation equalling 0 whereas you want to find a value of x for which \ln(3x^2+5x+3)&lt;0 as then the statement would be false if such real x value exists (a contradiction arises concerning the domain stated in the conjecture)

    To properly disprove it, think about \ln(a)\geq 0 and how a must be greater than or equal 1 for this to be true. So you want to show that \ln(a)&lt;0 exists therefore you want to solve a&lt;1 \Rightarrow 3x^2+5x+3&lt;1

 and show that there exists x values for which 3x^2+5x+3 is less than 1 - hence there exist x values for which \ln(3x^2+5x+3)&lt; 0

    P.S. If you have no knowledge of complex numbers then a real number is just any number you can think of.
    Oh okay so you basically just set up an inequality?

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    That looks almost right to me - a factor of 3 seems to have gone awol in the 3rd line (should be outside the square brackets), but reappears in the 4th. You shouldn't be setting it equal to zero, just rewriting y. The answer shows that the minimum value of y is less than 1. That point (calculate the x value) is your counter-example value.
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    (Original post by jessyjellytot14)
    Oh okay so you basically just set up an inequality?

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    Basically find a value of x for which \ln(3x^2+5x+3)&lt;0 - the inequality is just the key step which shows in what interval you can look for your x values once you solve it.
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    (Original post by RogerOxon)
    That looks almost right to me - a factor of 3 seems to have gone awol in the 3rd line (should be outside the square brackets), but reappears in the 4th. You shouldn't be setting it equal to zero, just rewriting y. The answer shows that the minimum value of y is less than 1. That point (calculate the x value) is your counter-example value.
    So then y = 3(x + 5/6)2 + 11/12 ?

    Which part of this would give me the x value? Or how do I calculate the x value from this?
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    In order to prove by counter-example that ln(3x^2 + 5x + 3) ≥ 0 for all x is false, you have to come up with a value of x (one will do) for which ln(3x^2 + 5x + 3)< 0, which is the same as 3x^2 + 5x + 3<1.

    Now, 3x^2 + 5x + 3 is a quadratic with positive "a" term, so it has a vase shape and therefore the turning point is a minimum. The turning point of any quadratic occurs at x=(-b/2a). If you can work out that value of x, and plug it into 3x^2 + 5x + 3, and if the result is <1, you have demonstrated that there is a value of x for which 3x^2 + 5x + 3<1, as needed.
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    (Original post by jessyjellytot14)
    So then y = 3(x + 5/6)2 + 11/12 ?

    Which part of this would give me the x value? Or how do I calculate the x value from this?
    Yes.

    Which value of x minimises y? Completing the square should make it easy to see, as the square of a real number is ..
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    (Original post by jessyjellytot14)
    So then y = 3(x + 5/6)2 + 11/12 ?

    Which part of this would give me the x value? Or how do I calculate the x value from this?
    If you look at a graph of y=ln(x), then you will see that it is negative for 0 < x < 1. So in your case, you need to find a value of x that makes 0 < 3x^2+5x+3 < 1. Now that you have completed the square, you can see that this is the same as needing 0 < 3(x + 5/6)2 + 11/12 < 1. Is there an obvious value of x that will do this for you? One that makes the 3(x + 5/6)2 + 11/12part much easier, by getting rid of part of it altogether?
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    (Original post by Pangol)
    If you look at a graph of y=ln(x), then you will see that it is negative for 0 < x < 1. So in your case, you need to find a value of x that makes 0 < 3x^2+5x+3 < 1. Now that you have completed the square, you can see that this is the same as needing 0 < 3(x + 5/6)2 + 11/12 < 1. Is there an obvious value of x that will do this for you? One that makes the 3(x + 5/6)2 + 11/12part much easier, by getting rid of part of it altogether?
    Ummm -5/6 ?
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    (Original post by jessyjellytot14)
    Ummm -5/6 ?
    That's the one!
 
 
 
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