Natural logarithmic equation help s'il vous plait

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.

"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

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...

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.

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 $\ln(3x^2+5x+3)<0$ 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 $\ln(a)<0$ exists therefore you want to solve $a<1 \Rightarrow 3x^2+5x+3<1[br]$ and show that there exists x values for which 3x^2+5x+3 is less than 1 - hence there exist x values for which $\ln(3x^2+5x+3)< 0$

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.

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?

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?

Ummm -5/6 ?