Exponential Inequalities: Difficult Problems with Solutions

Solution:
[tex]3^x\left(\sqrt{3}+\dfrac{1}{\sqrt{3}}\right)=\dfrac{4}{\sqrt{3}}\times 3^x, 2\times 4^x-\dfrac{1}{2}\times 4^x=\dfrac{3}{2}\times 4^x[/tex]
[tex]\left(\dfrac{3}{4}\right)^x\ge \dfrac{3\sqrt{3}}{8}=\left(\dfrac{3}{4}\right)^{\frac{3}{2}}[/tex]
The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]x\le \dfrac{3}{2}[/tex]

Solution:
[tex]x=1:\ 1\gt 1[/tex] — is false.
[tex]x\gt 1:\ 3x+1\gt x \iff x\gt -\dfrac{1}{2}\ \checkmark[/tex]
[tex]0\lt x\lt 1:\ 3x+1\lt x \iff x\lt -\dfrac{1}{2}[/tex] — no solution.
Answer: [tex]x\in(1, +\infty)[/tex]

Solution:
Substitute [tex]t=\left(\dfrac{1}{6}\right)^x\gt 0[/tex]: [tex]t^2-2t-35\lt 0 \iff (t-7)(t+5)\lt 0 \Rightarrow 0\lt t\lt 7[/tex]
[tex]6^{-x}\lt 7 \iff x\gt -\log_6 7, 1\lt \log_6 7\lt 2 \Rightarrow -2\lt -\log_6 7\lt -1[/tex]
Answer: [tex]-1[/tex]

Solution:
Substitute [tex]t=3^x\gt 0[/tex]: [tex]t^3-t^2-3t+3=(t-1)\left(t^2-3\right)=(t-1)\left(t-\sqrt{3}\right)\left(t+\sqrt{3}\right)\lt 0[/tex]
[tex]t+\sqrt{3}\gt 0 \Rightarrow 1\lt t\lt \sqrt{3} \iff 0\lt x\lt \dfrac{1}{2}[/tex]

Solution:
Dividing by [tex]9^x[/tex]; Substitute [tex]t=\left(\dfrac{2}{3}\right)^x\gt 0[/tex]: [tex]6t^2-13t+6=(3t-2)(2t-3)\gt 0 \Rightarrow t\lt \dfrac{2}{3}[/tex] or [tex]t\gt \dfrac{3}{2}[/tex]
The base is less than [tex]1[/tex], so the inequality sign reverses: [tex]x\gt 1[/tex] or [tex]x\lt -1[/tex]

Solution:
[tex]7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2[/tex]. Substitute [tex]t=\left(2-\sqrt{3}\right)^x\gt 0[/tex]: [tex]t^2-14t+1\le 0 \iff 7-4\sqrt{3}\le t\le 7+4\sqrt{3}[/tex]
[tex]7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2, 7+4\sqrt{3}=\left(2-\sqrt{3}\right)^{-2}[/tex]
[tex]0\lt 2-\sqrt{3}\lt 1 \Rightarrow -2\le x\le 2[/tex]
The integer solutions are [tex]-2, -1, 0, 1, 2[/tex].
Answer: [tex]5[/tex]

Solution:
[tex]x=1:\ a^{-3}\gt a^7[/tex]
For [tex]a=1[/tex] this is [tex]1\gt 1[/tex] (false); for [tex]a\gt 1[/tex] the function is increasing and [tex]a^{-3}\lt a^7[/tex]. So [tex]0\lt a\lt 1[/tex] and the function is decreasing:
[tex]x^2-4x\lt x+6 \iff x^2-5x-6\lt 0 \iff (x+1)(x-6)\lt 0 \iff -1\lt x\lt 6[/tex]

Solution:
Substitute [tex]t=3^x\gt 0, t\ne \dfrac{1}{3}[/tex]: [tex]\dfrac{3t^2-31t-19}{3t-1}+1\le 0 \iff \dfrac{3t^2-28t-20}{3t-1}\le 0 \iff \dfrac{(3t+2)(t-10)}{3t-1}\le 0[/tex]
[tex]3t+2\gt 0 \Rightarrow \dfrac{t-10}{3t-1}\le 0 \iff \dfrac{1}{3}\lt t\le 10 \iff -1\lt x\le \log_3 10[/tex]

Solution:
[tex]x^4\left(1-3^x\right)-81\left(1-3^x\right)=\left(1-3^x\right)(x-3)(x+3)\left(x^2+9\right)\ge 0[/tex]
[tex]1-3^x[/tex] is positive for [tex]x\lt 0[/tex] and negative for [tex]x\gt 0[/tex]; [tex](x-3)(x+3)[/tex] is positive for [tex]|x|\gt 3[/tex].
Answer: [tex]x\in(-\infty, -3]\cup[0, 3][/tex]

Solution:
[tex]7^x\ge 6^x+1 \iff \left(\dfrac{6}{7}\right)^x+\left(\dfrac{1}{7}\right)^x\le 1[/tex]
The left side [tex]g(x)[/tex] is strictly decreasing (both bases are less than [tex]1[/tex]) and [tex]g(1)=\dfrac{6}{7}+\dfrac{1}{7}=1[/tex], so [tex]g(x)\le g(1) \iff x\ge 1[/tex].

Solution:
Domain: [tex]x\ge \dfrac{1}{2}[/tex]. [tex]\sqrt{8x-4}=2\sqrt{2x-1}[/tex]:
[tex]4^{3\sqrt{2x-1}}\gt 8^x \iff 2^{6\sqrt{2x-1}}\gt 2^{3x} \iff 2\sqrt{2x-1}\gt x[/tex]
Squaring: [tex]8x-4\gt x^2 \iff x^2-8x+4\lt 0 \iff 4-2\sqrt{3}\lt x\lt 4+2\sqrt{3}[/tex]
[tex]4-2\sqrt{3}\approx 0. 54, 4+2\sqrt{3}\approx 7. 46[/tex]; The integer solutions are [tex]1, 2, \ldots, 7[/tex].
Answer: [tex]7[/tex]