Exponential Inequalities: Very Difficult Problems with Solutions

Solution:
[tex]10\times 3^{x^2}\le 10\times 5^{x^3} \iff 3^{x^2}\le 5^{x^3}[/tex]
Taking logarithms to base [tex]5[/tex]: [tex]x^2\log_5 3\le x^3 \iff x^2\left(x-\log_5 3\right)\ge 0[/tex]
[tex]x=0[/tex] or [tex]x\ge \log_5 3[/tex]

Solution:
[tex]3-x\gt 0 \iff x\lt 3[/tex]
[tex]x=2:\ 1\lt 1[/tex] — is false.
[tex]x\lt 2\ (3-x\gt 1):\ \dfrac{3x-5}{3-x}\lt 0 \iff 3x-5\lt 0 \iff x\lt \dfrac{5}{3}[/tex]
[tex]2\lt x\lt 3\ (0\lt 3-x\lt 1):\ \dfrac{3x-5}{3-x}\gt 0\ \checkmark[/tex]
Answer: [tex]x\in\left(-\infty, \dfrac{5}{3}\right)\cup(2, 3)[/tex]

Solution:
[tex]x\ne 0[/tex]
For [tex]x\lt 0[/tex]: [tex]3^x\lt 2^x[/tex], so the left side is negative and every [tex]x\lt 0[/tex] is a solution.
[tex]x\gt 0:\ 2\times 4^x\le 9^x-6^x[/tex]; Dividing by [tex]4^x[/tex], [tex]u=\left(\dfrac{3}{2}\right)^x\gt 1:\ u^2-u-2\ge 0 \iff (u-2)(u+1)\ge 0 \Rightarrow u\ge 2[/tex]
[tex]x\ge \log_{\frac{3}{2}}2[/tex]
Answer: [tex]x\in(-\infty, 0)\cup\left[\log_{\frac{3}{2}}2, +\infty\right)[/tex]

Solution:
Substitute [tex]t=3^x\gt 0[/tex]: [tex]\sqrt{3t^2-4t+1}\ge 3t-5[/tex]; Domain: [tex](3t-1)(t-1)\ge 0 \iff t\le \dfrac{1}{3}[/tex] or [tex]t\ge 1[/tex].
For [tex]t\lt \dfrac{5}{3}[/tex] the right side is negative, so the inequality holds on the whole domain:
[tex]t\le \dfrac{1}{3}[/tex] or [tex]1\le t\lt \dfrac{5}{3}[/tex]
[tex]t\ge \dfrac{5}{3}:\ 3t^2-4t+1\ge 9t^2-30t+25 \iff 3t^2-13t+12\le 0 \iff (3t-4)(t-3)\le 0 \Rightarrow \dfrac{5}{3}\le t\le 3[/tex]
[tex]t\le \dfrac{1}{3}[/tex] or [tex]1\le t\le 3 \iff x\le -1[/tex] or [tex]0\le x\le 1[/tex]

Solution:
Substitute [tex]t=2^x\gt 0[/tex]: [tex]1-t+\dfrac{t^2}{4}=\left(\dfrac{t}{2}-1\right)^2 \Rightarrow 8\sqrt{1-2^x+2^{2x-2}}=4|t-2|[/tex]; [tex]2^{2x}-2^{x+2}+7=(t-2)^2+3[/tex]
[tex]u=|t-2|\ge 0:\ u^2-4u+3\lt 0 \iff 1\lt u\lt 3[/tex]
[tex]1\lt |t-2|\lt 3, t\gt 0 \Rightarrow 0\lt t\lt 1[/tex] or [tex]3\lt t\lt 5[/tex]
[tex]x\lt 0[/tex] or [tex]\log_2 3\lt x\lt \log_2 5[/tex]

Solution:
Domain: [tex]1-2^{x+1}\ge 0 \iff x\le -1, x+2\gt 0, x\ne 0 \Rightarrow -2\lt x\le -1[/tex].
Substitute [tex]s=\sqrt{1-2^{x+1}}\in[0, 1), 2^x=\dfrac{1-s^2}{2}[/tex]: [tex]B=\dfrac{1+s-s^2}{2}, 0\le s-s^2\le \dfrac{1}{4} \Rightarrow \dfrac{1}{2}\le B\le \dfrac{5}{8}[/tex]
[tex]0\lt B\lt 1:\ B^E\gt 1 \iff E\lt 0 \iff \dfrac{3x^2}{8}\lt x+2 \iff (3x+4)(x-4)\lt 0 \iff -\dfrac{4}{3}\lt x\lt 4[/tex]
Together with the domain: [tex]x\in\left(-\dfrac{4}{3}, -1\right][/tex]

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]\left(\log_2 x-x\right)\left(5^x-4^x-3^x\right)\ge 0[/tex]
[tex]\log_2 x\lt x[/tex] for every [tex]x\gt 0[/tex] (because [tex]2^x\gt x[/tex]), so the first factor is negative and we need [tex]5^x-4^x-3^x\le 0[/tex]:
[tex]\left(\dfrac{4}{5}\right)^x+\left(\dfrac{3}{5}\right)^x\ge 1[/tex]
The left side is strictly decreasing and equals [tex]1[/tex] at [tex]x=2[/tex], so [tex]x\le 2[/tex].
Together with the domain: [tex]x\in(0, 2][/tex]

Solution:
Substitute [tex]t=2^{x^2}\ge 1[/tex]: [tex]at^2+4(a-1)t+a-1\gt 0 \iff a\left(t^2+4t+1\right)\gt 4t+1 \iff a\gt h(t)=\dfrac{4t+1}{t^2+4t+1}[/tex]
[tex]h'(t)=-\dfrac{4t^2+2t}{\left(t^2+4t+1\right)^2}\lt 0 \Rightarrow \max_{t\ge 1}h(t)=h(1)=\dfrac{5}{6}[/tex]
The inequality must hold for every [tex]t\ge 1[/tex], so [tex]a\gt \dfrac{5}{6}[/tex] (for [tex]a=\dfrac{5}{6}[/tex] the expression equals [tex]0[/tex] at [tex]x=0[/tex]).