Wyznaczamy dziedzinę: \[x\gt 0\]

Podstawiamy \(t=\log_5 x\): \[t^3-3t^2=t-3\\[6pt] t^3-3t^2-t+3=0\\[6pt] (t-1)(t+1)(t-3)=0\\[6pt] t=1 \quad \lor \quad t=-1 \quad \lor \quad t=3\]

Powrót do \(x\): \[ \begin{aligned} &t=1\\ &\log_5 x=1\\[6pt] &\boxed{x=5}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=-1\\ &\log_5 x=-1\\[6pt] &\boxed{x=\tfrac{1}{5}}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=3\\ &\log_5 x=3\\[6pt] &\boxed{x=125}_{\ \in \text{D}} \end{aligned} \]

Wyznaczamy dziedzinę: \[x\gt 0\]

Podstawiamy \(t=\ln x\): \[t^3+t^2-2t=0\\[6pt] t(t^2+t-2)=0\\[6pt] t=0 \quad \lor \quad t=-2 \quad \lor \quad t=1\]

Powrót do \(x\): \[ \begin{aligned} &t=0\\ &\ln x=0\\[6pt] &\boxed{x=1}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=-2\\ &\ln x=-2\\[6pt] &\boxed{x=e^{-2}}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=1\\ &\ln x=1\\[6pt] &\boxed{x=e}_{\ \in \text{D}} \end{aligned} \]

Wyznaczamy dziedzinę: \[x-1\gt 0\\[6pt] x\gt 1\]

Podstawiamy \(t=\log_2(x-1)\): \[t^4+2t^2=3\\[6pt] u:=t^2\\[6pt] u^2+2u-3=0\\[6pt] u=1 \quad \lor \quad u=-3\] Z \(u=1\) mamy \(t=1 \quad \lor \quad t=-1\).

Powrót do \(x\): \[ \begin{aligned} &t=1\\ &\log_2(x-1)=1\\[6pt] &x-1=2\\[6pt] &\boxed{x=3}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=-1\\ &\log_2(x-1)=-1\\[6pt] &x-1=\tfrac{1}{2}\\[6pt] &\boxed{x=\tfrac{3}{2}}_{\ \in \text{D}} \end{aligned} \]

Wyznaczamy dziedzinę: \[x\gt 0 \quad \land \quad \log x\ne 4 \quad \land \quad \log x\ne -2\]

Podstawiamy \(t=\log x\): \[\frac{1}{4-t}+\frac{2}{2+t}=1\\[6pt] \frac{10-t}{\,8+2t-t^2\,}=1\\[6pt] t^2-3t+2=0\\[6pt] t=1 \quad \lor \quad t=2\]

Powrót do \(x\): \[ \begin{aligned} &t=1\\ &\log x=1\\[6pt] &\boxed{x=10}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=2\\ &\log x=2\\[6pt] &\boxed{x=100}_{\ \in \text{D}} \end{aligned} \]

Wyznaczamy dziedzinę: \[x\gt 0 \quad \land \quad x\ne 1\]

Podstawiamy \(t=\ln^2 x\): \[t+\frac{3}{t}=4\\[6pt] t^2-4t+3=0\\[6pt] t=1 \quad \lor \quad t=3\]

Powrót do \(x\): \[ \begin{aligned} &t=1\\ &\ln^2 x=1\\[6pt] &\ln x=1 \quad \lor \quad \ln x=-1\\[6pt] &\boxed{x=e}_{\ \in \text{D}} \quad \lor \quad \boxed{x=e^{-1}}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &t=3\\ &\ln^2 x=3\\[6pt] &\ln x=\sqrt{3} \quad \lor \quad \ln x=-\sqrt{3}\\[6pt] &\boxed{x=e^{\sqrt{3}}}_{\ \in \text{D}} \quad \lor \quad \boxed{x=e^{-\sqrt{3}}}_{\ \in \text{D}} \end{aligned} \]

Wyznaczamy dziedzinę: \[x-2\gt 0 \quad \land \quad \log(x-2)\ne \pm 1\]

Podstawiamy \(t=\log(x-2)\): \[\frac{1+t}{1-t^2}+\frac{2}{1-t}=1\\[6pt] \frac{1}{1-t}+\frac{2}{1-t}=1\\[6pt] \frac{3}{1-t}=1\\[6pt] t=-2\]

Powrót do \(x\): \[ \begin{aligned} &t=-2\\ &\log(x-2)=-2\\[6pt] &x-2=10^{-2}\\[6pt] &\boxed{x=2{,}01}_{\ \in \text{D}} \end{aligned} \]