Logarithmic Equations: Very Difficult Problems with Solutions

Solution:
The equation is defined for [tex]1+x>0[/tex], [tex]1-x>0[/tex] and [tex]1-x^2>0[/tex]. The third one follows from the first two, so we can exclude it. What remains is [tex]\begin{array}{|l}x>-1\\x<1\end{array}[/tex], or [tex]x \in (-1;1)[/tex].
[tex]2+lg\sqrt{1+x}+3lg\sqrt{1-x}=lg\sqrt{1-x}+lg\sqrt{1+x}[/tex]
[tex]2+2lg\sqrt{1-x}=0[/tex]
[tex]lg\sqrt{1-x}=-1[/tex]
[tex]\sqrt{1-x}=10^{-1}=\frac{1}{10}[/tex]
[tex]1-x=\frac{1}{100}[/tex]

[tex]x=\frac{99}{100}[/tex]

Solution:
[tex]0. 04=\dfrac{1}{25}=5^{-2} \Rightarrow 0. 04^{-28}=5^{56}[/tex]
[tex]2+4+6+\ldots+2x=2(1+2+\ldots+x)=x(x+1)[/tex]
[tex]5^{x(x+1)}=5^{56} \iff x^2+x-56=0 \Rightarrow x=7\ \ (x=-8\notin\mathbb{N})[/tex]
Answer: [tex]x=7[/tex]

Solution:
Checking if the system is defined for two variables is a hard task, so we shall find the eventual solutions to the system and check directly if the system is defined for them. We shall only write [tex]\begin{array}{|l}x+y>0\\x-y>0\end{array}[/tex] for now.
[tex]\begin{array}{|l}\frac{x}{y}+\frac{y}{x}=log_432\\log_3(x^2-y^2)=1\end{array}[/tex]

[tex]\begin{array}{|l}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}log_232\\x^2-y^2=3\end{array}[/tex]

[tex]\begin{array}{|l}\frac{x^2+y^2}{xy}=\frac{5}{2}\\x^2-y^2=3\end{array}[/tex]

[tex]\begin{array}{|l}x^2+y^2=\frac{5}{2}xy\\x^2-y^2=3\end{array}[/tex]

[tex]\begin{array}{|l}x^2-2xy+y^2=\frac{1}{2}xy\\(x-y)(x+y)=3\end{array}[/tex]

[tex]\begin{array}{|l}(x-y)^2=\frac{xy}{2}\\(x-y)(x+y)=3\end{array}[/tex], we divide the first with the square of the second:

[tex]\frac{1}{(x+y)^2}=\frac{xy}{18}[/tex], or [tex](x+y)^2=\frac{18}{xy}[/tex]. We get the system
[tex]\begin{array}{|l}x^2+2xy+y^2=\frac{18}{xy}\\x^2-2xy+y^2=\frac{xy}{2}\end{array}[/tex], we now subtract them
[tex]4xy=\frac{18}{xy}-\frac{xy}{2}[/tex]
[tex]\frac{9}{2}xy=\frac{18}{xy}[/tex]
[tex](xy)^2=4[/tex]
[tex]xy = \pm 2[/tex]. But it can't be negative, because [tex]2(x-y)^2=xy[/tex], therefore [tex]xy=2[/tex]. Substituting back into the system, we get
[tex]\begin{array}{|l}(x-y)^2=x^2-2xy+y^2=x^2+y^2-4=1\\x^2-y^2=3\end{array}[/tex]

[tex]\begin{array}{|l}x^2+y^2=5\\x^2-y^2=3\end{array}[/tex], we subtract them:

[tex]2x^2=8[/tex], so [tex]x_1=2[/tex] and [tex]x_2=-2[/tex]. [tex]y_1=\frac{2}{x_1}=1[/tex] and [tex]y_2=\frac{2}{x_2}=-1[/tex]. We now have to check if they are solutions. [tex]x_1+y_1=3>0[/tex], [tex]x_1-y_1=1>0[/tex], so [tex](2;1)[/tex] is a solution to the system. [tex]x_2+y_2=-3<0[/tex], so [tex](-2;-1)[/tex] is not a solution.
The final solution is [tex]x=2[/tex].

Solution:
[tex]log_23 \times \frac{log_24}{log_23} \times \frac{log_25}{log_24} \cdot \cdot \cdot \frac{log_2(n+1)}{log_2n} = 10[/tex]
So [tex]log_{2}(n + 1) = 10[/tex]
[tex]=>n + 1 = 2^{10} => n=1023[/tex]

Solution:
Domain: [tex]x\gt 0[/tex].
Substitute [tex]\log_3 x=2t[/tex]: [tex]x=9^t, \sqrt{x}=3^t[/tex]
[tex]\log_2\left(1+3^t\right)=2t \iff 1+3^t=4^t \iff \left(\dfrac{1}{4}\right)^t+\left(\dfrac{3}{4}\right)^t=1[/tex]
The left side is strictly decreasing, so the equation has at most one root.
[tex]t=1:\ \dfrac{1}{4}+\dfrac{3}{4}=1\ \checkmark \Rightarrow x=9[/tex]
Check: [tex]\log_2 4=2=\log_3 9\ \checkmark[/tex]

Solution:
[tex]2\times 3^{x+1}-9^x-5=6\times 3^x-9^x-5=4-\left(3^x-3\right)^2\le 4 \Rightarrow \log_2\left(2\times 3^{x+1}-9^x-5\right)\le 2[/tex]
For [tex]x\lt 0[/tex]: [tex]0\lt 3^x\lt 1[/tex], so [tex]\left(3^x-3\right)^2\gt 4[/tex] and the argument is negative. For [tex]x\gt 0[/tex]: [tex]x+\dfrac{1}{x}\ge 2[/tex], with equality only for [tex]x=1[/tex]. So both sides must equal [tex]2[/tex], which gives [tex]x=1[/tex].
Check: [tex]\log_2(18-9-5)=\log_2 4=2=1+1\ \checkmark[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1[/tex]. [tex]\log_3 x^2=2\log_3 x, 2\log_x 9=\dfrac{4}{\log_3 x}, 1-2x+x^2=|x-1|^2[/tex]
Case [tex]|x-1|=1 \Rightarrow x=2[/tex] ([tex]x=0[/tex] is not in the domain); [tex]1=1\ \checkmark[/tex]
Case [tex]|x-1|\ne 1[/tex]: the exponents are equal: [tex]2\log_3 x-\dfrac{4}{\log_3 x}=2[/tex]
[tex]t=\log_3 x:\ t^2-t-2=0 \Rightarrow t=2, t=-1 \Rightarrow x=9, x=\dfrac{1}{3}[/tex]

Solution:
Domain: [tex]x\gt 1[/tex]. Substitute [tex]u=\log_2(x+1), v=\log_2(x-1)[/tex]: [tex]\log_4(x+1)=\dfrac{u}{2}, \log_2\left(x^2-1\right)=u+v[/tex]
Multiplying by [tex]6[/tex]: [tex]3(x+4)u-6(x-4)v=16u+16v \iff (3x-4)u-2(3x-4)v=0 \iff (3x-4)(u-2v)=0[/tex]
[tex]x=\dfrac{4}{3}\gt 1\ \checkmark[/tex]
[tex]u=2v \iff x+1=(x-1)^2 \iff x^2-3x=0 \Rightarrow x=3[/tex] ([tex]x=0[/tex] is not in the domain)

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]x=1:\ 0=0\ \checkmark[/tex]
[tex]x\ne 1[/tex]: Dividing by [tex]\log_2 x\times \log_3 x\times \log_5 x[/tex]:
[tex]1=\dfrac{1}{\log_5 x}+\dfrac{1}{\log_3 x}+\dfrac{1}{\log_2 x}=\log_x 5+\log_x 3+\log_x 2=\log_x 30 \Rightarrow x=30[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 2, x\ne \dfrac{1}{16}, x\ne \dfrac{1}{4}[/tex]. Substitute [tex]t=\log_2 x, t\ne 1, t\ne -4, t\ne -2[/tex]:
[tex]\log_{0. 5x}x^2=\dfrac{2t}{t-1}, \log_{16x}x^3=\dfrac{3t}{t+4}, \log_{4x}\sqrt{x}=\dfrac{t}{2(t+2)}[/tex]
[tex]\dfrac{2t}{t-1}-\dfrac{42t}{t+4}+\dfrac{20t}{t+2}=0[/tex]
[tex]t=0 \Rightarrow x=1[/tex]
[tex]t\ne 0:\ \dfrac{1}{t-1}-\dfrac{21}{t+4}+\dfrac{10}{t+2}=0 \iff -10t^2+15t+10=0 \iff 2t^2-3t-2=0[/tex]
[tex]t=2 \Rightarrow x=4, t=-\dfrac{1}{2} \Rightarrow x=2^{-\frac{1}{2}}=\dfrac{\sqrt{2}}{2}[/tex]

Solution:
Domain: [tex]x\gt 0, y\gt 0[/tex].
[tex]x^{\log_8 y}=8^{\log_8 x\times \log_8 y}=y^{\log_8 x}[/tex]
[tex]2x^{\log_8 y}=4 \Rightarrow x^{\log_8 y}=2 \Rightarrow \log_8 x\times \log_8 y=\dfrac{1}{3}[/tex]
[tex]\log_4\dfrac{x}{y}=1 \Rightarrow x=4y \Rightarrow \log_8 x=\log_8 y+\dfrac{2}{3}[/tex]
[tex]b=\log_8 y:\ b\left(b+\dfrac{2}{3}\right)=\dfrac{1}{3} \iff 3b^2+2b-1=0 \Rightarrow b=\dfrac{1}{3}, b=-1[/tex]
[tex]b=\dfrac{1}{3}:\ y=2, x=8[/tex]; [tex]b=-1:\ y=\dfrac{1}{8}, x=\dfrac{1}{2}[/tex]

Solution:
Domain: [tex]x\gt 0[/tex]. Substitute [tex]t=\log x, x=10^t, x^2=10^{2t}[/tex]; [tex]A=t^2-3, B=2t[/tex]:
[tex]\left(10^A-10^B\right)+10^t(A-B)=0[/tex]
Both terms have the same sign as [tex]A-B[/tex], so the sum is [tex]0[/tex] only when [tex]A=B[/tex].
[tex]t^2-2t-3=0 \Rightarrow t=3, t=-1 \Rightarrow x=1000, x=\dfrac{1}{10}[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1, x^2+x+a\ne 0[/tex].
[tex]\left(x^2+x+a\right)^2=x^4 \iff x^2+x+a=\pm x^2[/tex]
[tex]x^2+x+a=x^2 \Rightarrow x=-a, a\lt 0, a\ne -1[/tex]
[tex]x^2+x+a=-x^2 \iff 2x^2+x+a=0[/tex]
The sum of its roots is [tex]-\dfrac{1}{2}[/tex], so it has a positive root only for [tex]a\lt 0[/tex], and then exactly one: [tex]x_0=\dfrac{-1+\sqrt{1-8a}}{4}[/tex]; [tex]x_0=1[/tex] for [tex]a=-3[/tex]. The two candidates never coincide for [tex]a\lt 0[/tex]. Hence there is no solution for [tex]a\ge 0[/tex], two solutions for [tex]a\lt 0, a\ne -1, a\ne -3[/tex], and exactly one for [tex]a=-1[/tex] [tex]\left(x=\dfrac{1}{2}\right)[/tex] and for [tex]a=-3[/tex] [tex](x=3)[/tex].
Answer: [tex]a\in\{-3, -1\}[/tex]

Solution:
Domain: [tex]x\ne 0, \log_2 x^4\ge 0 \iff |x|\ge 1[/tex].
Substitute [tex]s=\sqrt{\log_2 x^4}\ge 0, \log_2 x^2=\dfrac{s^2}{2}[/tex]: [tex]s^2-2as-2a+2=0\ \ (*)[/tex]
Each root [tex]s\ge 0[/tex] gives exactly two values [tex]x=\pm 2^{\frac{s^2}{4}}[/tex], so we need two distinct roots [tex]s\ge 0[/tex] of [tex](*)[/tex]:
[tex]\dfrac{D}{4}=a^2+2a-2\gt 0, s_1+s_2=2a\gt 0, s_1s_2=2-2a\ge 0 \Rightarrow \sqrt{3}-1\lt a\le 1[/tex]
Answer: [tex]a\in\left(\sqrt{3}-1, 1\right][/tex]