For which value of does the following inequality hold?

Printable View

- March 13th 2008, 09:35 AMjames_bondLogarithm
For which value of does the following inequality hold?

- March 13th 2008, 10:04 AMMoo
Hello,

If there is no mistake in the subject, i'll write this :

So you have to solve :

Differentiate the case when ln(x²-3) is positive (ie x²-3 > 1) and the case when ln(x²-3) is negative (ie x²-3 < 1).

Multiply the two sides with ln(x²-3), change the sign when it's negative.

Then, you may have ln(4x+2) > (or <) ln(x²-3), which is ln(4x+2) - ln(x²-3) > (or <) 0=ln(1)

<=> (or <)

Ln is an increasing function, so a<b <=> ln(a)<ln(b) - March 13th 2008, 10:25 AMSoroban
Hello, james_bond!

This one is trickier than I thought . . .

Quote:

Solve for

We know that: . means: . (Think about it!)

So we have: .

This is a down-opening parabola which is positive between its x-intercepts.

Its x-intercepts are: (-1,0) and (5,0).

. . Hence, the inequality holds on the interval: .

But the base of a logarithm must be positive.

. . So we have: .

Moreover, the base of a logarithm cannot be 1.

. . So we have: .

Therefore: .

- March 13th 2008, 10:41 AMPlato
It is indeed tricky.

Recall that we need - March 13th 2008, 11:05 AMjames_bond
Hi Soroban!

I can't see a mistake in your solution but Wolfram Mathematica says that only is the solution. Am I wrong? (Worried)

EDIT: Yeah I didn't recognize Plato's post... Sorry.

I now think that the problem might be with the case...