Logarithmic Inequalities: Problems with Solutions

Solution:
Domain:
$\begin{array}{|l} 3-2x>0 \\ 4x + 1 > 0 \end{array} \Leftrightarrow \begin{array}{|l} 3>2x \\ 4x > -1 \end{array} \Leftrightarrow \begin{array}{|l}x < \frac{2}{3} \\ x > -\frac{1}{4} \end{array} \Leftrightarrow x \in \left( -\frac14, \frac23\right) $

$\log_5(3-2x) \ge \log_5(4x+1)$
$3-2x \ge 4x+1$
$2 \ge 6x$
$1 \ge 3x$
$x \le \frac13$, but

Domain: $x \in \left( -\frac14; \frac23 \right) \Rightarrow x \in \left(-\frac14; \frac13\right]$

Answer: $x \in \left(-\frac14; \frac13\right]$

Solution:
Domain:
$\begin{array}{|l} 10x + 3 > 0 \\ 7x -4> 0 \end{array} \Leftrightarrow \begin{array}{|l} 10x > -3 \\ 7x > 4 \end{array} \Leftrightarrow \begin{array}{|l} x > -\frac{3}{10} \\ x > \frac47 \end{array} \Leftrightarrow x > \frac47$

$\log_{0.3}(10x + 3) < \log_{0.3}(7x - 4)$

$0.3 < 1 \Rightarrow$

$10x + 3 > 7x - 4$

$3x > -7$
$x > -\frac73$, but

Domain: $x > \frac47$
$\Rightarrow x \in \left(\frac47, +\infin \right)$

Solution:
Domain:

$\begin{array}{|l} x > 0 \\ x + 1 > 0 \\ 2x > 0 \end{array} \Leftrightarrow$ $\begin{array}{|l} x > 0 \\ x > -1 \\ x > 0 \end{array}$ $\Leftrightarrow x > 0 \Leftrightarrow x \in (0; \infin)$

$\lg x + \lg (x+1) > \lg(2x)$

$\lg[x(x+1)] > \lg(2x)$

$\lg(x^2+x) > \lg(2x)$

$x^2+x > 2x$

$x^2-x > 0$

$x(x-1) > 0$

$x\in(-\infin, 0)\cup(1, +\infin)$ but

Domain: $x\in(0, +\infin) \Rightarrow x\in(1, +\infin)$

Answer: $x\in(1,+\infin)$

Solution:
Domain:

$\begin{array}{|l} x > 0 \\ 2-x > 0 \end{array} \Leftrightarrow \begin{array}{|l} x > 0 \\ x < 2 \end{array} \Leftrightarrow x \in (0, 2)$

$\lg x + \lg(2 - x) < 1$

$\lg[x \cdot(2 - x)] < \lg10$

$\lg(2x - x^2) < \lg10$

$2x - x^2 < 10$

$x^2 - 2x + 10 > 0$

$x^2 - 2x + 10 = 0$

$D = -36 < 0 \Rightarrow x^2 - 2x + 10 > 0$ - for every x

But domain is: $x \in (0, 2)$

Answer $x \in (0, 2)$

Solution:
Domain:

$\begin{array}{|l} x > 0 \\ x + 1 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x > 0 \\ x > -1 \end{array} \Leftrightarrow x \in (0, +\infin)$

$\lg x - \lg(x + 1) > 0$

$\lg x > \lg(x + 1)$

$x > x + 1$

$0\cdot x > 1$

$x \in \varnothing$

Answer: $x \in \varnothing$

Solution:
Domain:

$\begin{array}{|l} 3x-1 > 0 \\ x + 2 > 0 \end{array} \Leftrightarrow \begin{array}{|l} 3x > 1 \\ x > -2 \end{array} \Leftrightarrow \begin{array}{|l} x > \frac13 \\ x > -2 \end{array} \Leftrightarrow x \in (\frac13, +\infin)$

$\log_{\frac13}(3x-1)-\log_{\frac13}(x+2) > 1$

$\log_{\frac13} \frac{3x-1}{x+2} > \log_{\frac13}\frac13$

$\frac{3x-1}{x+2} < \frac13$

$\frac{3x-1}{x+2} - \frac13$<0

$\frac{3(3x-1) - (x+2)}{3(x+2)}$<0

$\frac{9x - 3- x-2}{3(x+2)}<0$

$\frac{8x - 5}{3(x+2)}<0$     $/\times 3$

$\frac{8x - 5}{x+2}<0$

Domain $x\ne2$

$(8x - 5)(x+2)<0 \Leftrightarrow x\in(-2, \frac58)$

Domain: $x \in (\frac13, +\infin)$

Answer: $x \in \left(\frac13, \frac58\right)$

Solution:
Domain:
$\begin{array}{|l} 2x^2+3x+1>0 \\ 2x + 2 > 0 \end{array}$

$2x^2+3x+1>0$

$D=1$

$x_1=-1,\ x_2=-\frac12$

$x\in(-\infin,-1)\cup(-\frac12,+\infin)$

$\begin{array}{|l} 2x^2+3x+1>0 \\ 2x + 2 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x\in(-\infin,-1)\cup(-\frac12,+\infin) \\ x > -1 \end{array} \Leftrightarrow x \in \left( -\frac12, +\infin\right) $

$\log_4(2x^2+3x+1)\le \log_2(2x+2)$

$\log_{2^2}(2x^2+3x+1)\le \log_2(2x+2)$

$\frac12\log_2(2x^2+3x+1)\le \log_2(2x+2)$ $\ \ \ \ /\times2$

$\log_2(2x^2+3x+1)\le 2 \log_2(2x+2)$

$\log_2(2x^2+3x+1)\le \log_2(2x+2)^2$

$2x^2+3x+1\le (2x+2)^2$

$2x^2+3x+1\le 4x^2+8x+4$

$-2x^2-5x-3\le 0\ \ \ \ /\times-1$

$2x^2+5x+3\ge 0$

$D=1$

$x_1=-\frac32, \ x_2=-1$

$x\in \left(-\infin, -\frac32\right]\cup \left[-1,+\infin\right)$

Domain: $x \in \left( -\frac12, +\infin\right) \Rightarrow$

Answer: $x \in \left( -\frac12, +\infin\right)$

Solution:
Domain:
$\begin{array}{|l} x^2-x-6>0 \\ x-3 > 0 \end{array}$

$x^2-x-6=0$

$D=25$

$x_1=-2, \ x_2=3$

$x^2-x-6=(x+2)(x-3)$

$x^2-x-6>0 \Leftrightarrow (x+2)(x-3)>0 \Leftrightarrow x\in(-\infin, -2)\cup(3, +\infin)$


$\begin{array}{|l} x^2-x-6>0 \\ x-3 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x\in(-\infin, -2)\cup(3, +\infin) \\ x \in(3, +\infin) \end{array} \Leftrightarrow x \in(3, +\infin)$

$\log_2(x+2)(x-3)+\log_{0.5}(x-3)<2\log_23$

$\log_2(x+2)(x-3)+\log_{2^{-1}}(x-3)<\log_23^2$

$\log_2(x+2)(x-3)-\log_{2}(x-3)<\log_29$

$\log_2\frac{(x+2)(x-3)}{x-3}<\log_29$

$\log_2x+2<\log_29$

$x+2<9$

$x<7 \ x \in (-\infin, 7)$

Domain: $x\in(-\infin, -2)\cup(3, +\infin)$

Answer: $x \in (3, 7)$

Solution:
Domain:
$\begin{array}{|l} 2x^2+x+1>0 \\ 2x - 1 > 0 \end{array}$

$2x^2+x+1 > 0$

$2x^2+x+1 = 0$

$D=-7 < 0$

$\Rightarrow 2x^2+x+1 > 0 - \forall x$


$\begin{array}{|l} 2x^2+x+1>0 \\ 2x - 1 > 0 \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\ x > \frac12 \end{array} \Leftrightarrow x \in \left( \frac12, +\infin\right)$

$\log_4(2x^2+x+1)-\log_2(2x-1)\le 1$

$\log_{2^2}(2x^2+x+1)-\log_2(2x-1)\le 1$

$\frac12\log_2(2x^2+x+1)-\log_2(2x-1)\le 1 \ \ \ /\times2$

$\log_2(2x^2+x+1)-2\log_2(2x-1)\le 2$

$\log_2(2x^2+x+1)-\log_2(2x-1)^2\le 2$

$\log_2\frac{2x^2+x+1}{(2x-1)^2}\le 2\times 1$

$\log_2\frac{2x^2+x+1}{(2x-1)^2}\le 2\times \log_22$

$\log_2\frac{2x^2+x+1}{(2x-1)^2}\le \log_22^2$

$\frac{2x^2+x+1}{(2x-1)^2}\le 2^2$

$\frac{2x^2+x+1}{(2x-1)^2} - 4\le 0$

$\frac{2x^2+x+1-4(2x-1)^2}{(2x-1)^2}\le 0$

$\frac{2x^2+x+1-4(4x^2-4x+1)}{(2x-1)^2}\le 0$

$\frac{2x^2+x+1-16x^2+16x-4}{(2x-1)^2}\le 0$

$\frac{-14x^2+17x-3}{(2x-1)^2}\le 0$

$(2x-1)^2 > 0$

$\Rightarrow \frac{-14x^2+17x-3}{(2x-1)^2}\le 0 \Leftrightarrow -14x^2+17x-3 \le 0$

$-14x^2+17x-3 \le 0 \ \ \ /\times-1$

$14x^2-17x+3 \ge 0$

$D=121$

$x_1=1, \ \ x_2=\frac{3}{14}$

$14x^2-17x+3 \ge 0 \Leftrightarrow 14(x-1)(x-\frac{3}{14})\ge 0 \Leftrightarrow x \in (-\infin, \frac{3}{14}) \cup (1, +\infin)$

Domain: $x \in \left( -\frac12, +\infin \right)$

Answer: $x \in (1, +\infin)$

Solution:
Domain:
$\begin{array}{|l} x+1>0 \\ x-1 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x>-1 \\ x>1 \end{array} \Leftrightarrow x\in(1, +\infin)$

$\log_{3^{\frac{1}{2}}}(x+1)-\log_{3^{\frac{1}{2}}}(x-1)>\log_34$

$\frac{1}{\frac{1}{2}}\log_3(x+1)-\frac{1}{\frac{1}{2}}\log_3(x-1)>\log_32^2$

$2\log_3(x+1)-2\log_3(x-1)>\log_32^2$

$2\log_3(x+1)-2\log_3(x-1)>2\log_32 \ \ \ \ /\div 2$

$\log_3(x+1)-\log_3(x-1)>\log_32$

$\log_3\frac{x+1}{x-1}>\log_32$

$\frac{x+1}{x-1}>2$

$\frac{x+1}{x-1}-2>0$

$\frac{(x+1)-2(x-1)}{x-1}>0$

$\frac{x+1-2x-2}{x-1}>0$

$\frac{-x+3}{x-1}>0$

$\frac{-(x-3)}{x-1}>0 \ \ \ \ /\times(-1)$

$\frac{x-3}{x-1}<0 \Leftrightarrow x \in (1, 3)$

Answer: $x \in (1, 3)$

Solution:
Domain:
$\begin{array}{|l} x>0 \\ x^2-3x+2 > 0 \end{array}$

$x^2-3x+2 > 0$

$D=1$

$x_1=1,\ \ x_2=2$

$x^2-3x+2 > 0 \Leftrightarrow (x-1)(x-2)>0 \Leftrightarrow x\in (-\infin, 1)\cup (2, +\infin)$


$\begin{array}{|l} x>0 \\ x^2-3x+2 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x\in(0, \infin) \\ x\in (-\infin, 1)\cup (2, +\infin) \end{array} \Leftrightarrow x\in (-\infin, 1)\cup (2, +\infin)$

$2\log_4x - 0.5\log_2(x^2-3x+2)\le -0.5$

$2\log_{2^2}x - \frac12\log_2(x^2-3x+2)\le -\frac12$

$\frac{2}{2}\log_2x - \frac12\log_2(x^2-3x+2)\le -\frac12$

$\log_2x - \frac12\log_2(x^2-3x+2)\le -\frac12 \ \ \ \ \ /\times 2$

$2\log_2x - \log_2(x^2-3x+2)\le -1$

$\log_2x^2 - \log_2(x^2-3x+2)\le -1$

$\log_2\frac{x^2}{x^2-3x+2}\le -1$

$\log_2\frac{x^2}{x^2-3x+2}\le -1\times 1$

$\log_2\frac{x^2}{x^2-3x+2}\le -1\times \log_22$

$\log_2\frac{x^2}{x^2-3x+2}\le \log_22^{-1}$

$\log_2\frac{x^2}{x^2-3x+2}\le \log_2\frac12$

$\frac{x^2}{x^2-3x+2}\le \frac12$

$\frac{x^2}{x^2-3x+2} - \frac12 \le 0$

$\frac{2x^2-(x^2-3x+2)}{2(x^2-3x+2)} \le 0$

$\frac{2x^2-x^2+3x-2}{2(x^2-3x+2)} \le 0$

$\frac{x^2+3x-2}{2(x^2-3x+2)} \le 0$

Domain: $x^2-3x+2 > 0 \Leftrightarrow \frac{x^2+3x-2}{2(x^2-3x+2)} \le 0 \Leftrightarrow x^2+3x-2 \le 0$

$x^2+3x-2 \le 0$

$D=17$

$x_1,2=\frac{-3\pm\sqrt{17}}{2}$

$x_1=-\frac{3+\sqrt{17}}{2}\ \ \ \ \ x_2=\frac{\sqrt{17}-3}{2}$

$x^2+3x-2 \le 0 \Leftrightarrow (x+\frac{3+\sqrt{17}}{2})(x-\frac{\sqrt{17}-3}{2})\le 0 \Leftrightarrow x\in \left[-\frac{3+\sqrt{17}}{2}, \frac{\sqrt{17}-3}{2}\right]$

Domain: $x\in (-\infin, 1)\cup (2, +\infin)$

$\Rightarrow x\in \left(0, \frac{\sqrt{17}-3}{2}\right]$

Solution:
Domain: $x>0$

$|3-\log_2x|<2$

$\begin{array}{|l} 3-\log_2x<2 \\ 3-\log_2x>-2 \end{array}$

$\begin{array}{|l} -\log_2x<2-3 \\ -\log_2x>-2-3 \end{array}$

$\begin{array}{|l} -\log_2x<-1 \\ -\log_2x>-5 \end{array} $

$\begin{array}{|l} \log_2x>1 \\ \log_2x<5 \end{array} $

$\begin{array}{|l} \log_2x>\log_22 \\ \log_2x<\log_22^5 \end{array}$

$\begin{array}{|l} x>2 \\ x<32 \end{array} \Leftrightarrow x \in (2, 32)$

Solution:
Domain:
$\begin{array}{|l} x > 0 \\ \log_2x -1 \ne 0 \end{array} \Leftrightarrow \begin{array}{|l} x> 0 \\ \log_2x \ne 1 \end{array} \Leftrightarrow \begin{array}{|l} x > 0 \\ \log_2x \ne \log_22 \end{array} \begin{array}{|l} x > 0 \\ x \ne 2 \end{array} \Rightarrow x \in (0, 2)\cup (2, +\infin)$

Let $\log_2x=u$

$\frac{u^2-3u+3}{u-1}>1, \ \ u\ne 1$

$\frac{u^2-3u+3}{u-1} - 1 > 0$

$\frac{u^2-3u+3-(u-1)}{u-1} > 0$

$\frac{u^2-3u+3-u+1}{u-1} > 0$

$\frac{u^2-4u+4}{u-1} > 0$

$\frac{(u-2)^2}{u-1} > 0 \Leftrightarrow (u-2)^2\cdot(u-1) > 0$

$(u-2)^2>0\ \ \ - \forall u\ne 2 \Rightarrow$

$(u-2)^2\cdot(u-1) > 0$

$\begin{array}{|l} u-1 > 0 \\ u\ne 2 \end{array} \Leftrightarrow \begin{array}{|l} u > 1\\ u \ne 2 \end{array} \Leftrightarrow \begin{array}{|l} \log_2x > 1 \\ \log_2x \ne 2 \end{array}$

$\Leftrightarrow \begin{array}{|l} \log_2x > \log_22 \\ \log_2x \ne \log_24 \end{array} \Leftrightarrow \begin{array}{|l} x > 2 \\ x \ne 4 \end{array} \Rightarrow\\ x \in (2, 4)\cup (4, +\infin)$

Answer: $x \in (2, 4)\cup (4, +\infin)$

Solution:
Domain:
$\begin{array}{|l} x^2 - 6x+8 > 0 \\ x-8 > 0 \\ \log_2(x-8) \ne 0 \end{array}$

$x^2 - 6x+8 = 0$

$D=4$

$x_1=2,\ \ x_2=4$

$x^2 - 6x+8 > 0 \Leftrightarrow (x-2)(x-4)>0$

$\Leftrightarrow x\in(-\infin, 2)\cup(4, +\infin)\Rightarrow$

$\begin{array}{|l} x^2 - 6x+8 > 0 \\ x-8 > 0 \\ \log_2(x-8) \ne 0 \end{array} \Leftrightarrow \begin{array}{|l} x\in(-\infin, 2)\cup(4, +\infin) \\ x> 8 \\ \log_2(x-8) \ne \log_21 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (8, +\infin) \\ x-8 \ne 1 \end{array} \Rightarrow x \in (8, 9)\cup (9, +\infin)$


$\frac{\log_2(x^2-6x+8)}{\log_2(x-8)}<1$

$\frac{\log_2(x^2-6x+8)}{\log_2(x-8)}-1<0$

$\frac{\log_2(x^2-6x+8)-\log_2(x-8)}{\log_2(x-8)}<0$

Case 1:

$\begin{array}{|l} \log_2(x^2-6x+8)-\log_2(x-8) > 0 \\ \log_2(x-8) < 0\end{array} \Leftrightarrow$

$\begin{array}{|l} \log_2(x^2-6x+8)>\log_2(x-8) \\ \log_2(x-8) < \log_21\end{array} \Leftrightarrow$

$\begin{array}{|l} x^2-6x+8>x-8 \\ x-8 < 1\end{array} \Leftrightarrow$

$\begin{array}{|l} x^2-7x+16>0 \\ x < 9\end{array} \Leftrightarrow \begin{array}{|l} \forall x(D< 0) \\ x < 9\end{array} \Rightarrow x \in (-\infin, 9),$ but

$x \in (8, 9)\cup (9, +\infin) \Rightarrow x \in (8,9)$


OR

Case 2:

$\begin{array}{|l} \log_2(x^2-6x+8)-\log_2(x-8) < 0 \\ \log_2(x-8) > 0\end{array} \Leftrightarrow$

$\begin{array}{|l} \log_2(x^2-6x+8)<\log_2(x-8) \\ \log_2(x-8) > \log_21\end{array} \Leftrightarrow$

$\begin{array}{|l} x^2-6x+8 < x-8 \\ x-8 > 1\end{array} \Leftrightarrow$

$\begin{array}{|l} x^2-7x+16 < 0 \\ x > 9\end{array} \Leftrightarrow \begin{array}{|l} x \in \varnothing(D< 0) \\ x < 9\end{array} \Rightarrow x \in \varnothing$

Answer: $x \in (8, 9)$

Solution:
Domain:
$\begin{array}{|l} x+1 \ne 0 \\ \frac{2x+1}{x+1} > 0 \\ \log_2\frac{2x+1}{x+1} > 0\end{array} \Leftrightarrow $

$\begin{array}{|l} x \ne -1 \\ (2x+1)(x+1) > 0 \\ \log_2\frac{2x+1}{x+1} > \log_21\end{array} \Leftrightarrow $

$\begin{array}{|l} x \ne -1 \\ x\in(-\infin,-1)\cup(-\frac12, +\infin) \\ \frac{2x+1}{x+1} > 1\end{array} \Leftrightarrow $

$\begin{array}{|l} x\in(-\infin,-1)\cup(-\frac12, +\infin) \\ \frac{2x+1}{x+1} -1> 0\end{array} \Leftrightarrow $

$\begin{array}{|l} x\in(-\infin,-1)\cup(-\frac12, +\infin) \\ \frac{2x+1-x-1}{x+1}> 0\end{array} \Leftrightarrow $

$\begin{array}{|l} x\in(-\infin,-1)\cup(-\frac12, +\infin) \\ x(x+1)> 0\end{array} \Leftrightarrow $

$\begin{array}{|l} x\in(-\infin,-1)\cup(-\frac12, +\infin) \\ x\in(-\infin, -1)\cup(0,+\infin)\end{array} \Leftrightarrow x\in(-\infin,-1)\cup(0, +\infin)$


$\log_{0.5}\left(\log_2\frac{2x+1}{x+1}\right)>0$

$\log_{0.5}\left(\log_2\frac{2x+1}{x+1}\right)>\log_{0.5}1$

$\log_2\frac{2x+1}{x+1}<1$

$\log_2\frac{2x+1}{x+1}<\log_22$

$\frac{2x+1}{x+1}<2$

$\frac{2x+1}{x+1}-2 < 0$

$\frac{2x+1-2x-2}{x+1} < 0$

$\frac{-1}{x+1} < 0$

$\frac{1}{x+1} > 0$

$x+1 > 0 \Rightarrow x > -1$

Domain: $x\in(-\infin,-1)\cup(0, +\infin)$

Answer: $x\in(0, +\infin)$