For which value of $\displaystyle x$ does the following inequality hold?
$\displaystyle \log_{x^2-3}\left(4x+2\right)\ge1$
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)
Hello, james_bond!
This one is trickier than I thought . . .
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: . $\displaystyle (\sqrt{3},\,2) \:\cup \:(2,\,5)$
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?
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...