Logarithmic Equations: Very Difficult Problems with Solutions

Solution:
The equation is defined for 1+x>0, 1-x>0 and 1-x^2>0. 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 = \frac{1}{25}[/tex]
[tex]5^{2(1+2+3+....+x)}=\left(\frac{1}{25}\right)^{-28}[/tex]
[tex]1+2+3+....+x[/tex] is an arithmetic series and it is equal to
[tex]1+2+3+....+x = \frac{x(x + 1)}{2}[/tex]
[tex]5^x(x+1)=5^56[/tex]
[tex]x(x+1)=56[/tex]
[tex]x^2+x-56 = 0[/tex] - a quadratic equation with a positive root x = 7

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 x=2.

Solution:
[tex]log_23.\frac{log_24}{log_23}.\frac{log_25}{log_24} \cdot \cdot \cdot \frac{log_2(n+1)}{log_2n} = 10[/tex]
So log₂(n + 1) = 10
=>n + 1 = 2¹⁰ => n=1023