1
$\begingroup$

From the following inequality: $\dfrac{\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x)}{\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geq 0 $, the set of all real solutions to this inequality is of the form: $$(a) \ (a,b) \cup (b,c) \\ (b) \ (-\infty,a) \cup (c,\infty) \\ (c) \ (a,b) $$ for some real numbers $a,b,c $ such that $-\infty<a<b<c<+\infty $.

First I found the points at which the numerator and the denominator equal $0$. I found that from the following: $$\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x) = 0 \\ \log_{2x^2 + 2x + 3}(x^2 - 2x) = 1 $$ $$2x^2 + 2x + 3 = x^2 - 2x .$$

From this quadratic equation i find that the zeros of the function are $x_1=-3 $ and $x_2=-1$ and let them be $a=-1 $ and $b=-3 $ for now. I also found from the plot of the given quadratic equation that $(-\infty,-3)\cup(-1,\infty) $ makes function a positive one and the values in between $(-3,-1)$ make it negative. I used that later for the rational number chart.

Similarly $$\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0 \\ x^2 + 6x + 10 = 1 \\ x = -3.$$ This function is positive for every other value. Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative I get the positive values to range from: $(-\infty,-3) \cup (-1,\infty)$ which then makes my answer of the form $(-\infty,b) \cup (a, +\infty)$.

First of all, it's not given, second, it's wrong. The solution I have is for the $(a)$ answer. I just don't know how to get to it.

$\endgroup$
2
  • $\begingroup$ Just a comment on formatting: you realize that single dollar signs place your math inline and not display mode like double dollar signs do, right? $\endgroup$ – Xoque55 Dec 9 '15 at 17:46
  • $\begingroup$ Did not know that. Thanks :) $\endgroup$ – diredragon Dec 9 '15 at 19:44
1
$\begingroup$

Do you know about the change of base formula? If $a,b,x > 0$, $a \not= 1$, and $x \not= 1$ then $$ \log_a b = \frac{\log_x b}{\log_x a}.$$ Your original inequality reads $$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$$ provided that $2^{x^2 + 2x + 1} - 1 > 0$, $2^{x^2 + 2x + 1} - 1 \not= 1$, $x^2 + 6x + 10 > 0$, $x^2 + 6x + 10 \not= 1$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) > 0$, and $2x^2 + 2x + 3 > 0$. These are all true provided that $x \not= 0$, $x \not= -2$, $x \not= -1$, $x \not= -3$, and $x^2 - 2x > 1$.

Next observe $$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0 \iff \log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1 \\ \iff x^2 - 2x \ge (2x^2 + 2x + 3)^1.$$

The last inequality reduces to $x^2 + 4x + 3 \le 0$ which has the solution $[-3,-1]$. Note that $x^2 - 2x \ge 3$ on this interval, so excluding $x = -1$, $x = -2$, and $x = -3$ you get $$ (-3,-2) \cup (-2,-1).$$

$\endgroup$
2
  • $\begingroup$ You need $2^{x^2+2x+1}-1\not=1$. $\endgroup$ – mathlove Dec 9 '15 at 18:19
  • $\begingroup$ Good eye! Thanks! $\endgroup$ – Umberto P. Dec 9 '15 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.