Logarithmic Inequality Problems: Difficult Problems with Solutions

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]\log_{0. 3}\dfrac{3x}{10}=\log_{0. 3}(0. 3x)=1+\log_{0. 3}x[/tex]
[tex]4\log_{0. 3}x\le 3+\log_{0. 3}x \iff \log_{0. 3}x\le 1=\log_{0. 3}0. 3[/tex]
The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]x\ge 0. 3[/tex]
Answer: [tex]0. 5[/tex]

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]\log_{\frac{1}{2}}x=-\log_2 x \Rightarrow \left(\log_{\frac{1}{2}}x\right)^2=\left(\log_2 x\right)^2[/tex]
Substitute [tex]t=\log_2 x[/tex]: [tex]t^2\le 3t \iff 0\le t\le 3 \iff 1\le x\le 8[/tex]
The integer solutions are [tex]1, 2, \ldots, 8[/tex].
Answer: [tex]8[/tex]

Solution:
Domain: [tex]3x-x^2\gt 0, 8x-x^2\gt 0 \Rightarrow 0\lt x\lt 3[/tex].
[tex]\log_3\dfrac{1}{8x-x^2}=-\log_3\left(8x-x^2\right)=\log_{\frac{1}{3}}\left(8x-x^2\right)[/tex]
The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]3x-x^2\le 8x-x^2 \iff x\ge 0[/tex]
Together with the domain: [tex]x\in(0, 3)[/tex]

Solution:
Domain: [tex]x\gt 0, 1+\log_{\frac{1}{9}}x-\log_9 x\gt 0[/tex].
[tex]\log_{\frac{1}{9}}x=-\log_9 x \Rightarrow 1+\log_{\frac{1}{9}}x-\log_9 x=1-2\log_9 x=1-\log_3 x[/tex]
[tex]\log_2\left(1-\log_3 x\right)\lt 1=\log_2 2 \iff 0\lt 1-\log_3 x\lt 2[/tex]
[tex]-1\lt \log_3 x\lt 1 \iff \dfrac{1}{3}\lt x\lt 3[/tex]

Solution:
Domain: [tex]x\ne -1, x\ne \dfrac{1}{2}[/tex].
[tex]0\lt \left|\dfrac{2x-1}{x+1}\right|\lt 1 \iff |2x-1|\lt |x+1|, x\ne \dfrac{1}{2}[/tex]
[tex](2x-1)^2\lt (x+1)^2 \iff 3x^2-6x\lt 0 \iff 0\lt x\lt 2[/tex]
Answer: [tex]x\in\left(0, \dfrac{1}{2}\right)\cup\left(\dfrac{1}{2}, 2\right)[/tex]

Solution:
Domain: [tex]x^2-x\gt 0 \iff x\lt 0[/tex] or [tex]x\gt 1[/tex].
[tex]\log_4\left(16x^2\right)=2+\log_4 x^2=2+\log_2|x|[/tex]
[tex]\log_2\left[2\left(x^2-x\right)\right]\le \log_2(4|x|) \iff x^2-x\le 2|x|[/tex]
[tex]x\gt 1:\ x^2-x\le 2x \iff x\le 3 \Rightarrow 1\lt x\le 3[/tex]
[tex]x\lt 0:\ x^2-x\le -2x \iff x^2+x\le 0 \iff -1\le x\le 0 \Rightarrow -1\le x\lt 0[/tex]

Solution:
Domain: [tex]2^x\lt 12[/tex].
[tex]\log_2\left(12-2^x\right)\ge 5-x \iff 12-2^x\ge 2^{5-x}=\dfrac{32}{2^x}[/tex]
Substitute [tex]t=2^x, 0\lt t\lt 12[/tex]: [tex]12t-t^2\ge 32 \iff t^2-12t+32\le 0 \iff 4\le t\le 8[/tex]
[tex]2\le x\le 3[/tex]

Solution:
Domain: [tex]x\ge -1, \sqrt{x+1}-x\gt 0[/tex].
[tex]2=\log_{\frac{1}{2}}\dfrac{1}{4}[/tex]. The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]\sqrt{x+1}-x\gt \dfrac{1}{4}[/tex]
Substitute [tex]s=\sqrt{x+1}\ge 0, x=s^2-1[/tex]: [tex]s-s^2+1\gt \dfrac{1}{4} \iff 4s^2-4s-3\lt 0 \iff (2s-3)(2s+1)\lt 0 \Rightarrow 0\le s\lt \dfrac{3}{2}[/tex]
[tex]0\le x+1\lt \dfrac{9}{4} \iff -1\le x\lt \dfrac{5}{4}[/tex]

Solution:
[tex]\log_{\frac{1}{6}}\dfrac{x+1}{6x-1}=\log_6\dfrac{6x-1}{x+1}=A, A\gt 0[/tex]
[tex]\log_5 A\lt -\log_5 A \iff \log_5 A\lt 0 \iff 0\lt A\lt 1 \iff 1\lt \dfrac{6x-1}{x+1}\lt 6[/tex]
[tex]\dfrac{6x-1}{x+1}\gt 1 \iff \dfrac{5x-2}{x+1}\gt 0 \iff x\lt -1[/tex] or [tex]x\gt \dfrac{2}{5}[/tex]
[tex]\dfrac{6x-1}{x+1}\lt 6 \iff \dfrac{-7}{x+1}\lt 0 \iff x\gt -1[/tex]
Answer: [tex]x\in\left(\dfrac{2}{5}, +\infty\right)[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1, \dfrac{4x-6}{x-1}\gt 0 \Rightarrow x\in(0, 1)\cup\left(\dfrac{3}{2}, +\infty\right)[/tex]
Case [tex]x\gt \dfrac{3}{2}:\ \dfrac{4x-6}{x-1}\lt x \iff 4x-6\lt x^2-x \iff x^2-5x+6\gt 0 \Rightarrow x\in\left(\dfrac{3}{2}, 2\right)\cup(3, +\infty)[/tex]
Case [tex]0\lt x\lt 1:\ \dfrac{4x-6}{x-1}\gt x \iff 4x-6\lt x^2-x \iff x^2-5x+6\gt 0\ \checkmark[/tex]
Answer: [tex]x\in(0, 1)\cup\left(\dfrac{3}{2}, 2\right)\cup(3, +\infty)[/tex]

Solution:
Domain: [tex]|x+1|\gt 0, |x+1|\ne 1 \iff x\ne -1, x\ne 0, x\ne -2[/tex].
[tex]\log_{|x+1|}\left[(x+1)^4\right]=\log_{|x+1|}|x+1|^4=4[/tex]
[tex]16+\log_2(x+1)^2\le 22 \iff (x+1)^2\le 64 \iff -9\le x\le 7[/tex]
The interval [tex][-9, 7][/tex] contains [tex]17[/tex] integers; removing [tex]-2, -1, 0[/tex] leaves [tex]14[/tex].
Answer: [tex]14[/tex]

Solution:
Domain: [tex]9-3^x\gt 0 \iff x\lt 2[/tex].
[tex]0\lt 9-3^x\lt 9 \Rightarrow \log_3\left(9-3^x\right)-3\lt -1\lt 0[/tex]
Multiplying by the negative denominator reverses the sign:
[tex]x-1\ge \log_3\left(9-3^x\right)-3 \iff \log_3\left(9-3^x\right)\le x+2 \iff 9-3^x\le 9\times 3^x \iff 3^x\ge \dfrac{9}{10}[/tex]
[tex]x\ge \log_3\dfrac{9}{10}[/tex]. Together with the domain: [tex]x\in\left[\log_3\dfrac{9}{10}, 2\right)[/tex]

Solution:
Domain: [tex]x\gt 0, 2x^2-7x+6=(2x-3)(x-2)\gt 0, 2x^2-7x+6\ne 1[/tex]
[tex]x\in(0, 1)\cup\left(1, \dfrac{3}{2}\right)\cup\left(2, \dfrac{5}{2}\right)\cup\left(\dfrac{5}{2}, +\infty\right)[/tex]
[tex]2x^2-7x+6\gt 1 \iff (x-1)(2x-5)\gt 0:\ \dfrac{x}{3}\gt 1 \iff x\gt 3 \Rightarrow x\in(3, +\infty)[/tex]
[tex]0\lt 2x^2-7x+6\lt 1\ \left(x\in\left(1, \dfrac{3}{2}\right)\cup\left(2, \dfrac{5}{2}\right)\right):\ \dfrac{x}{3}\lt 1\ \checkmark[/tex]
Answer: [tex]x\in\left(1, \dfrac{3}{2}\right)\cup\left(2, \dfrac{5}{2}\right)\cup(3, +\infty)[/tex]