Logarithmic Inequality Problems: Problems with Solutions

Solution:
[tex]\sqrt{20}\lt \sqrt{25}=5[/tex]
The base is greater than [tex]1[/tex], so [tex]\log_2\sqrt{20}\lt \log_2 5[/tex].

Solution:
[tex]\dfrac{1}{2}\log_4 5=\log_4\sqrt{5}, \sqrt{5}\lt \sqrt{9}=3[/tex]
The base is greater than [tex]1[/tex], so [tex]\log_4\sqrt{5}\lt \log_4 3[/tex].
The statement is false.

Solution:
Domain: [tex]x-1\gt 0 \iff x\gt 1[/tex].
[tex]2=\log_4 16[/tex]. The base is greater than [tex]1[/tex], so [tex]x-1\le 16 \iff x\le 17[/tex].
Together with the domain: [tex]x\in(1, 17][/tex]

Solution:
Domain: [tex]x\gt 1[/tex].
[tex]2=\log_{0. 2}0. 04[/tex]. The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]x-1\lt 0. 04=\dfrac{1}{25} \iff x\lt \dfrac{26}{25}[/tex]
Together with the domain: [tex]x\in\left(1, \dfrac{26}{25}\right)[/tex]

Solution:
[tex]a\gt 0, a\ne 1[/tex]
Since [tex]\dfrac{1}{3}\gt \dfrac{1}{4}[/tex] and the larger number has the larger logarithm, the function [tex]\log_a t[/tex] is increasing, i.e. [tex]a\gt 1[/tex]. (For [tex]0\lt a\lt 1[/tex] it is decreasing and [tex]\log_a\dfrac{1}{3}\lt \log_a\dfrac{1}{4}[/tex].)

Solution:
Domain: [tex]x\ge 0, 3-\sqrt{x}\gt 0 \iff 0\le x\lt 9[/tex].
[tex]0=\log_{\frac{1}{3}}1[/tex]. The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]3-\sqrt{x}\ge 1 \iff \sqrt{x}\le 2 \iff x\le 4[/tex]
Together with the domain: [tex]x\in[0, 4][/tex]

Solution:
Domain: [tex]x^2\gt 3. 5[/tex].
[tex]1=\log_{0. 5}0. 5[/tex]. The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]x^2-3. 5\ge 0. 5 \iff x^2\ge 4[/tex]
[tex]x\le -2[/tex] or [tex]x\ge 2[/tex] [tex](x^2\ge 4\gt 3. 5\ \checkmark)[/tex]

Solution:
Domain: [tex]5x+10\gt 0, (x+2)(x+4)\gt 0 \Rightarrow x\gt -2[/tex].
The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]5x+10\gt x^2+6x+8 \iff x^2+x-2\lt 0 \iff (x+2)(x-1)\lt 0 \iff -2\lt x\lt 1[/tex]
Answer: [tex]x\in(-2, 1)[/tex]

Solution:
[tex]x^2-8x+16=(x-4)^2[/tex]; Domain: [tex]x\ne 4[/tex].
[tex]0=\log_{0. 3}1[/tex]. The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]0\lt (x-4)^2\lt 1 \iff |x-4|\lt 1, x\ne 4[/tex]
[tex]3\lt x\lt 5, x\ne 4[/tex]

Solution:
[tex]x^2-x+2\gt 0[/tex] for every [tex]x\ \ (D=-7\lt 0)[/tex].
[tex]0=\log_2 1[/tex]. The base is greater than [tex]1[/tex], so [tex]x^2-x+2\le 1 \iff x^2-x+1\le 0[/tex].
[tex]x^2-x+1\gt 0[/tex] for every [tex]x\ \ (D=-3\lt 0)[/tex] — no solution.

Solution:
Domain: [tex]x+2\gt 0, 2x-6\gt 0 \iff x\gt 3[/tex].
[tex]\log \left[(x+2)(2x-6)\right]\lt \log 100 \iff (x+2)(2x-6)\lt 100 \iff x^2-x-56\lt 0 \iff -7\lt x\lt 8[/tex]
Together with the domain: [tex]3\lt x\lt 8[/tex]. The integer solutions are [tex]4, 5, 6, 7[/tex].
Answer: [tex]4[/tex]