Wyznaczamy dziedzinę: \[x\gt 0\]
Niech \(t=\log x\). Własność \(x^{A}=10^{A\log x}=10^{At}\): \[10^{(t^3-\tfrac{1}{2}t)\,t}=10^{\tfrac{1}{2}}\\[6pt] t^4-\tfrac{1}{2}t^2=\tfrac{1}{2}\\[6pt] 2t^4-t^2-1=0\\[6pt] u:=t^2\\[6pt] 2u^2-u-1=0\\[6pt] \Delta=1+8=9\\[6pt] u=1\ \lor\ u=-\tfrac{1}{2}\] Bierzemy \(u=t^2=1\), więc \(t=1\ \lor\ t=-1\).
Powrót do \(x\): \[ \begin{aligned} &\log x=1\\[6pt] &\boxed{x=10}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &\log x=-1\\[6pt] &\boxed{x=0{,}1}_{\ \in \text{D}} \end{aligned} \]
Wyznaczamy dziedzinę: \[x\gt 0\]
Niech \(t=\log x\), wtedy: \[10^{t(2-t)}=10^{1}\\[6pt] t(2-t)=1\\[6pt] -t^2+2t-1=0\\[6pt] (t-1)^2=0\\[6pt] t=1\]
Powrót do \(x\): \[ \begin{aligned} &\log x=1\\[6pt] &\boxed{x=10}_{\ \in \text{D}} \end{aligned} \]
Wyznaczamy dziedzinę: \[ \begin{split} x\gt 0\ \land\ 4x\gt 0 \end{split} \] Zatem: \(x\gt 0\).
Niech \(t=\log_4 x\). Wówczas \(\log_4(4x)=1+t\) oraz \(x=4^{t}\): \[ (4^{t})^{\,1+t}=16\\[6pt] 4^{\,t(1+t)}=4^{2}\\[6pt] t(1+t)=2\\[6pt] t^2+t-2=0\\[6pt] \Delta=1+8=9,\ \ t=1\ \lor\ t=-2 \]
Powrót do \(x\): \[ \begin{aligned} &\log_4 x=1\\[6pt] &\boxed{x=4}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &\log_4 x=-2\\[6pt] &\boxed{x=\tfrac{1}{16}}_{\ \in \text{D}} \end{aligned} \]
Wyznaczamy dziedzinę: \[x\gt 0\]
Niech \(t=\log_7 x\). Ponieważ \(\sqrt{x}=7^{t/2}\): \[ \big(7^{t/2}\big)^{\,t-1}=7\\[6pt] 7^{\,\frac{t}{2}(t-1)}=7^{1}\\[6pt] \frac{t}{2}(t-1)=1\\[6pt] t^2-t-2=0\\[6pt] \Delta=1+8=9,\ \ t=2\ \lor\ t=-1 \]
Powrót do \(x\): \[ \begin{aligned} &\log_7 x=2\\[6pt] &\boxed{x=49}_{\ \in \text{D}} \end{aligned} \quad \lor \quad \begin{aligned} &\log_7 x=-1\\[6pt] &\boxed{x=\tfrac{1}{7}}_{\ \in \text{D}} \end{aligned} \]