Logarithmic Inequality Problems: Very Difficult Problems with Solutions

Solution:
[tex]\dfrac{2^{x+1}-5\times 3^x}{2^x-3^{x+1}}-1=\dfrac{2^x-2\times 3^x}{2^x-3\times 3^x}\lt 0[/tex]
[tex]x_0=\dfrac{1}{\log 32}\gt 0 \Rightarrow 2^{x_0}\lt 3^{x_0} \Rightarrow 2^{x_0}-2\times 3^{x_0}\lt 0, 2^{x_0}-3\times 3^{x_0}\lt 0[/tex]
The fraction is positive, so this number is not a solution.

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]x^{\log_5 x}=\left(5^{\log_5 x}\right)^{\log_5 x}=5^{\log_5^2 x}[/tex]
[tex]2\times 5^{\log_5^2 x}\lt 10 \iff 5^{\log_5^2 x}\lt 5 \iff \log_5^2 x\lt 1 \iff -1\lt \log_5 x\lt 1 \iff \dfrac{1}{5}\lt x\lt 5[/tex]

Solution:
Domain: [tex]-2\lt x\lt 9[/tex].
[tex]f(x)=\log_3(x+2)[/tex] is increasing and [tex]g(x)=\log_2(18-2x)[/tex] is decreasing, so [tex]f(x)-g(x)[/tex] is increasing and has at most one zero.
[tex]x=7:\ \log_3 9=2=\log_2 4[/tex]
[tex]f(x)\gt g(x) \iff x\gt 7[/tex]. Together with the domain: [tex]x\in(7, 9)[/tex]

Solution:
Domain: [tex]0\lt x\lt \log_2 6, x\ne 1[/tex].
[tex]\log_x 2=\dfrac{1}{\log_2 x}:\ \dfrac{\log_2\left(6-2^x\right)}{\log_2 x}\le 1[/tex]
Case [tex]1\lt x\lt \log_2 6\ \left(\log_2 x\gt 0\right):\ \log_2\left(6-2^x\right)\le \log_2 x \iff 2^x+x\ge 6[/tex]
[tex]h(x)=2^x+x[/tex] is strictly increasing, [tex]h(2)=6 \Rightarrow 2\le x\lt \log_2 6[/tex]
Case [tex]0\lt x\lt 1\ \left(\log_2 x\lt 0\right):\ 2^x+x\le 6[/tex]; [tex]h(x)\lt h(1)=3\ \checkmark[/tex]
Answer: [tex]x\in(0, 1)\cup[2, \log_2 6)[/tex]

Solution:
[tex]p\gt 0, p\ne 1[/tex]. [tex]-\log_{\frac{1}{p}}10=\log_p 10, -\log_{\frac{1}{p}}2=\log_p 2[/tex]
[tex]\log_p\left[10\left(2\times 3^x+1\right)\right]\le \log_p\left[2\left(9^x+14\right)\right][/tex]
Substitute [tex]t=3^x\gt 0[/tex]: [tex]2\left(t^2+14\right)-10(2t+1)=2t^2-20t+18=2(t-1)(t-9)[/tex]
[tex]p\gt 1:\ (t-1)(t-9)\ge 0 \Rightarrow x\in(-\infty, 0]\cup[2, +\infty)[/tex]
[tex]0\lt p\lt 1:\ (t-1)(t-9)\le 0 \Rightarrow 1\le t\le 9 \Rightarrow x\in[0, 2][/tex]

Solution:
Domain: [tex]0\lt x\lt 2, x\ne 1[/tex].
[tex]\log_4(2-x)-\log_{14}(2-x)=\ln(2-x)\left(\dfrac{1}{\ln 4}-\dfrac{1}{\ln 14}\right)=\ln(2-x)\times \dfrac{\ln\frac{7}{2}}{\ln 4\times \ln 14}[/tex]
[tex]\log_{14}x-\log_{49}x=\ln x\times \dfrac{\ln\frac{7}{2}}{\ln 14\times \ln 49}[/tex]
[tex]\dfrac{\ln(2-x)}{\ln x}\times \dfrac{\ln 49}{\ln 4}=\log_x(2-x)\times \log_4 49\le \log_4 49 \iff \log_x(2-x)\le 1[/tex]
[tex]1\lt x\lt 2:\ 2-x\le x \iff x\ge 1\ \checkmark[/tex]
[tex]0\lt x\lt 1:\ 2-x\ge x \iff x\le 1\ \checkmark[/tex]
Answer: [tex]x\in(0, 1)\cup(1, 2)[/tex]

Solution:
Domain: [tex]x\gt 0[/tex]. [tex]t=\log_2 x:\ t^2+(x-2)t-(x-1)=(t-1)(t+x-1)\le 0[/tex]
[tex]\left(\log_2 x-1\right)\left(\log_2 x+x-1\right)\le 0[/tex]
[tex]\log_2 x+x-1[/tex] is increasing and vanishes at [tex]x=1[/tex]; [tex]\log_2 x-1[/tex] vanishes at [tex]x=2[/tex]. For [tex]0\lt x\lt 1[/tex] both factors are negative, for [tex]x\gt 2[/tex] both are positive, so the solution set is [tex][1, 2][/tex].
[tex]u=2^x\in[2, 4]:\ 2^x+2^{2-x}=u+\dfrac{4}{u}[/tex] is strictly increasing [tex](u\ge 2)[/tex]
[tex]\max\left(u+\dfrac{4}{u}\right)=4+1=5 \Rightarrow a\ge 5[/tex]

Solution:
Substitute [tex]A=x^2-4x-11[/tex]. Domain: [tex]A\gt 0 \iff x\lt 2-\sqrt{15}[/tex] or [tex]x\gt 2+\sqrt{15}[/tex]; [tex]2-5x-3x^2=-(3x-1)(x+2)\ne 0 \Rightarrow x\ne -2[/tex].
[tex]2\log_5 A-3\log_{11}A=\ln A\times \dfrac{2\ln 11-3\ln 5}{\ln 5\times \ln 11}=\ln A\times \dfrac{\ln 121-\ln 125}{\ln 5\times \ln 11}[/tex]
The constant factor is negative ([tex]121\lt 125[/tex]), so the numerator has the sign of [tex]1-A[/tex], where [tex]A-1=x^2-4x-12=(x-6)(x+2)[/tex].
[tex]\dfrac{-(x-6)(x+2)}{-(3x-1)(x+2)}\ge 0 \iff \dfrac{x-6}{3x-1}\ge 0, x\ne -2 \iff x\lt \dfrac{1}{3}[/tex] or [tex]x\ge 6[/tex]
Together with the domain: [tex]x\in(-\infty, -2)\cup\left(-2, 2-\sqrt{15}\right)\cup[6, +\infty)[/tex]

Solution:
[tex]4x-x^2-3=1-(x-2)^2\le 1, 1\le 1+\cos^2\pi x\le 2 \Rightarrow 0\le \log_2\left(1+\cos^2\pi x\right)\le 1[/tex]
If [tex]4x-x^2-3\le 0[/tex], the product is [tex]\le 0\lt 1[/tex]. Otherwise the product is at most [tex]1\times 1=1[/tex], with equality only when [tex](x-2)^2=0[/tex] and [tex]\cos^2\pi x=1[/tex].
[tex]x=2:\ \cos 2\pi=1 \Rightarrow 1\times 1=1\ \checkmark[/tex]
Answer: [tex]x=2[/tex]

Solution:
Domain: [tex]x\gt 0, (x-1)(x+2)\gt 0 \Rightarrow x\gt 1[/tex].
[tex]2\log_{\frac{8}{9}}x=-\log_{\frac{9}{8}}x^2 \Rightarrow \log_{\frac{9}{8}}\dfrac{x^2+x-2}{x^2}=\log_{\frac{9}{8}}\left(1+\dfrac{1}{x}-\dfrac{2}{x^2}\right)[/tex]
[tex]u=\dfrac{1}{x}\in(0, 1):\ 1+u-2u^2=\dfrac{9}{8}-2\left(u-\dfrac{1}{4}\right)^2\le \dfrac{9}{8}[/tex]
[tex]\dfrac{1}{2}x^4-4x^3+8x^2+1=\dfrac{1}{2}x^2(x-4)^2+1\ge 1[/tex]
So the left side is at most [tex]1[/tex] and the right side is at least [tex]1[/tex]; in both cases equality holds only for [tex]x=4[/tex].
Answer: [tex]x=4[/tex]

Solution:
[tex]a\gt 0, a\ne 1[/tex]. Substitute [tex]u=x^2-ax+5\ge 0[/tex]:
[tex]\dfrac{\ln 3-F(u)}{\ln a}\ge 0, F(u)=\ln\left(1+\sqrt{u}\right)\times \log_5(u+1)[/tex]
[tex]F[/tex] is strictly increasing [tex](u\ge 0), F(4)=\ln 3\times \log_5 5=\ln 3[/tex]
[tex]0\lt a\lt 1:\ F(u)\ge \ln 3 \iff u\ge 4 \iff x^2-ax+1\ge 0[/tex] — infinitely many solutions.
[tex]a\gt 1:\ 0\le u\le 4 \iff x^2-ax+1\le 0, x^2-ax+5\ge 0[/tex]
For [tex]1\lt a\lt 2[/tex] there are no solutions; for [tex]a=2[/tex]: [tex](x-1)^2\le 0[/tex], [tex]x=1[/tex], and [tex]1-2+5=4\ge 0[/tex] — exactly one solution; for [tex]a\gt 2[/tex] a whole interval of [tex]x[/tex] works.
Answer: [tex]a=2[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1, x\ne 1-a[/tex].
[tex]p=\log_2(x+a), q=\log_2 x:\ \dfrac{1}{p}\lt \dfrac{2}{q}[/tex]
[tex]x\gt 1\ (p\gt 0, q\gt 0):\ q\lt 2p \iff x\lt (x+a)^2\ \checkmark[/tex], [tex](x+a)^2\gt x^2\gt x[/tex]
For [tex]1-a\lt x\lt 1[/tex]: [tex]p\gt 0\gt q[/tex], the left side is positive and the right side negative — no solutions.
[tex]0\lt x\lt 1-a\ (p\lt 0, q\lt 0):\ q\lt 2p \iff x^2+(2a-1)x+a^2\gt 0, D=1-4a\gt 0[/tex]
[tex]x_{1,2}=\dfrac{1-2a\mp\sqrt{1-4a}}{2}, 0\lt x_1\lt x_2\lt 1-a \Rightarrow 0\lt x\lt x_1[/tex] or [tex]x_2\lt x\lt 1-a[/tex]