Logarithmic Equations: Very Difficult Problems with Solutions

Solution:
Domain: [tex]x\gt 0[/tex]. Taking logarithms to base [tex]5[/tex]; [tex]c=\log_5 3, t=\log_5 x[/tex]: [tex]\log_5 15=c+1, \log_5(9x)=2c+t[/tex]
[tex]c(c+1)+(2c+t+1)t=0 \iff t^2+(2c+1)t+c(c+1)=0 \iff (t+c)(t+c+1)=0[/tex]
[tex]t=-c=-\log_5 3 \Rightarrow x=\dfrac{1}{3}, t=-c-1=-\log_5 15 \Rightarrow x=\dfrac{1}{15}[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1[/tex]. [tex]t=\log_2 x=\dfrac{1}{\log_x 2}[/tex]: [tex]t^2+(x-1)t+2x-6=0[/tex]
[tex]D=(x-1)^2-4(2x-6)=x^2-10x+25=(x-5)^2[/tex]
[tex]t=\dfrac{1-x\pm(x-5)}{2}:\ t=-2, t=3-x[/tex]
[tex]\log_2 x=-2 \Rightarrow x=\dfrac{1}{4}[/tex]
[tex]\log_2 x=3-x[/tex]: The left side is strictly increasing and the right side is strictly decreasing, so the equation has at most one root. [tex]x=2:\ 1=1\ \checkmark[/tex]

Solution:
For every admissible [tex]a[/tex], [tex]\log_a x[/tex] takes all real values, so [tex]t=\log_a^2 x[/tex] takes all values in [tex][0, +\infty)[/tex]. We need [tex]f(t)=4t^2-4mt+8m-7=0[/tex] to have a root [tex]t\ge 0[/tex].
[tex]f(0)=8m-7\le 0 \iff m\le \dfrac{7}{8}:\ t_1t_2\le 0\ \checkmark[/tex]
[tex]m\gt \dfrac{7}{8}:\ t_1t_2\gt 0, t_1+t_2=m\gt 0, \dfrac{D}{4}=4m^2-4(8m-7)=4(m-1)(m-7)\ge 0[/tex]
[tex]\dfrac{7}{8}\lt m\le 1[/tex] or [tex]m\ge 7[/tex]
Answer: [tex]m\in(-\infty, 1]\cup[7, +\infty)[/tex]

Solution:
Substitute [tex]t=\log x[/tex]: [tex]a^2-\left(2t^2+1\right)a+t^4+t=0, D=(2t-1)^2[/tex]
[tex]a=t^2+t[/tex] or [tex]a=t^2-t+1 \iff \left(t^2+t-a\right)\left(t^2-t+1-a\right)=0[/tex]
[tex]t^2+t-a=0:\ 1+4a\gt 0 \iff a\gt -\dfrac{1}{4}[/tex]
[tex]t^2-t+1-a=0:\ 4a-3\gt 0 \iff a\gt \dfrac{3}{4}[/tex]
Each real [tex]t[/tex] gives exactly one [tex]x\gt 0[/tex]. A common root of the two factors requires [tex]2t-1=0[/tex], i.e. [tex]t=\dfrac{1}{2}[/tex], which happens only for [tex]a=\dfrac{3}{4}[/tex]. So there are four distinct roots exactly when [tex]a\gt \dfrac{3}{4}[/tex].
Answer: [tex]a\in\left(\dfrac{3}{4}, +\infty\right)[/tex]