# Logarithm

• March 13th 2008, 08:35 AM
james_bond
Logarithm
For which value of $x$ does the following inequality hold?
$\log_{x^2-3}\left(4x+2\right)\ge1$
• March 13th 2008, 09:04 AM
Moo
Hello,

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

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

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

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)

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

Ln is an increasing function, so a<b <=> ln(a)<ln(b)
• March 13th 2008, 09:25 AM
Soroban
Hello, james_bond!

This one is trickier than I thought . . .

Quote:

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

We know that: . $\log_b(N) \:\geq \:1$ means: $N \:\geq \:b\quad\hdots$ .

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

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

But the base of a logarithm must be positive.
. . So we have: . $x^2-3 \:>\:0\quad\Rightarrow\quad |x| \:>\: \sqrt{3}$

Moreover, the base of a logarithm cannot be 1.
. . So we have: . $x^2-3 \:\neq \:1\quad\Rightarrow\quad x \neq \pm 2$

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

• March 13th 2008, 09:41 AM
Plato
It is indeed tricky.
$1.8 \in \left( {\sqrt 3 ,2} \right)\quad \Rightarrow \quad \log _{1.8} \left( {4(1.8) + 2} \right) = - 1.555$
Recall that we need $x^2 - 3 > 1$
• March 13th 2008, 10:05 AM
james_bond
Hi Soroban!

I can't see a mistake in your solution but Wolfram Mathematica says that only $(2,5)$ 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 $0 case...