For which value of $\displaystyle x$ does the following inequality hold?

$\displaystyle \log_{x^2-3}\left(4x+2\right)\ge1$

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- Mar 13th 2008, 08:35 AMjames_bondLogarithm
For which value of $\displaystyle x$ does the following inequality hold?

$\displaystyle \log_{x^2-3}\left(4x+2\right)\ge1$ - Mar 13th 2008, 09:04 AMMoo
Hello,

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

$\displaystyle \log_{x} y = \frac{\ln(y)}{ln(x)}$

So you have to solve : $\displaystyle \frac{\ln(4x+2)}{\ln(x^2 -3)} \geq 1$

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)

<=> $\displaystyle \ln(\frac{4x+2}{x^2-3}) \geq ln(1)$ (or <)

Ln is an increasing function, so a<b <=> ln(a)<ln(b) - Mar 13th 2008, 09:25 AMSoroban
Hello, james_bond!

This one is trickier than I thought . . .

Quote:

Solve for $\displaystyle x\!:\;\;\log_{x^2-3}(4x+2)\:\ge\:1$

We know that: .$\displaystyle \log_b(N) \:\geq \:1$ means: $\displaystyle N \:\geq \:b\quad\hdots$ . (Think about it!)

So we have: . $\displaystyle 4x + 2 \:\geq \:x^2-3 \quad\Rightarrow\quad 5 + 4x - x^2 \:\geq \:0$

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: .$\displaystyle (-1,\,5)$

But the base of a logarithm must be positive.

. . So we have: .$\displaystyle x^2-3 \:>\:0\quad\Rightarrow\quad |x| \:>\: \sqrt{3}$

Moreover, the base of a logarithm cannot be 1.

. . So we have: .$\displaystyle x^2-3 \:\neq \:1\quad\Rightarrow\quad x \neq \pm 2$

Therefore: . $\displaystyle (\sqrt{3},\,2) \:\cup \:(2,\,5)$

- Mar 13th 2008, 09:41 AMPlato
It is indeed tricky.

$\displaystyle 1.8 \in \left( {\sqrt 3 ,2} \right)\quad \Rightarrow \quad \log _{1.8} \left( {4(1.8) + 2} \right) = - 1.555$

Recall that we need $\displaystyle x^2 - 3 > 1$ - Mar 13th 2008, 10:05 AMjames_bond
Hi Soroban!

I can't see a mistake in your solution but Wolfram Mathematica says that only $\displaystyle (2,5)$ 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 $\displaystyle 0<x^2-3<1$ case...