Wyznaczamy dziedzinę: \[ \begin{split} x-1\gt 0\ &\land\ 2x-10\gt 0\\[6pt] x\gt 1\ &\land\ x\gt 5 \end{split} \] Zatem: \[ x\gt 5 \]
Podnosimy do potęgi i odejmujemy logarytmy: \[ \log_3\!\big((x-1)^2\big)-\log_3 9=\log_3(2x-10)\\[6pt] \log_3\!\left(\frac{(x-1)^2}{9}\right)=\log_3(2x-10)\\[6pt] \frac{(x-1)^2}{9}=2x-10\\[6pt] x^2-2x+1=18x-90\\[6pt] x^2-20x+91=0\\[6pt] \Delta=20^2-4\cdot 1\cdot 91=400-364=36 \] Zatem: \[ \boxed{x=7}_{\ \in \text{D}} \quad \lor \quad \boxed{x=13}_{\ \in \text{D}} \]
Wyznaczamy dziedzinę: \[ \begin{split} 5x-4\gt 0\ &\land\ x+1\gt 0\\[6pt] x\gt \tfrac{4}{5}\ &\land\ x\gt -1 \end{split} \] Zatem: \[ x\gt \tfrac{4}{5} \]
Zamieniamy pierwiastki na potęgi i dodajemy logarytmy: \[ \tfrac{1}{2}\log(5x-4)+\tfrac{1}{2}\log(x+1)=2+\log 0{,}18\\[6pt] \tfrac{1}{2}\log\!\big((5x-4)(x+1)\big)=\log 18\\[6pt] \log\!\big((5x-4)(x+1)\big)=\log 324\\[6pt] (5x-4)(x+1)=324\\[6pt] 5x^{2}+x-328=0\\[6pt] \Delta=1^2-4\cdot 5\cdot(-328)=1+6560=6561 \] Zatem: \[ \boxed{x=8}_{\ \in \text{D}} \quad \lor \quad x=-\tfrac{41}{5}_{\ \notin \text{D}} \]
Wyznaczamy dziedzinę: \[ \begin{split} x-5\gt 0\ &\land\ 2x-3\gt 0\\[6pt] x\gt 5\ &\land\ x\gt \tfrac{3}{2} \end{split} \] Zatem: \[ x\gt 5 \]
Zamieniamy pierwiastek na potęgę, dodajemy logarytmy i przenosimy \(1=\log 10\): \[ \tfrac{1}{2}\log(x-5)+\tfrac{1}{2}\log(2x-3)=\log 30-\log 10\\[6pt] \tfrac{1}{2}\log\!\big((x-5)(2x-3)\big)=\log 3\\[6pt] \log\!\big((x-5)(2x-3)\big)=\log 9\\[6pt] (x-5)(2x-3)=9\\[6pt] 2x^{2}-13x+6=0\\[6pt] \Delta=(-13)^2-4\cdot 2\cdot 6=169-48=121 \] Zatem: \[ \boxed{x=6}_{\ \in \text{D}} \quad \lor \quad x=\tfrac{1}{2}_{\ \notin \text{D}} \]
Wyznaczamy dziedzinę: \[ \begin{split} x\gt 0\ &\land\ x+4\gt 0 \end{split} \] Zatem: \[ x\gt 0 \]
Zamieniamy pierwiastek na potęgę i dodajemy logarytmy: \[ \tfrac{1}{2}\log_4 x+\tfrac{1}{2}\log_4(x+4)=\tfrac{5}{4}\\[6pt] \log_4\!\big(x(x+4)\big)=\tfrac{5}{2}\\[6pt] x(x+4)=4^{5/2}\\[6pt] x^{2}+4x-32=0\\[6pt] \Delta=4^2-4\cdot 1\cdot(-32)=16+128=144 \] Zatem: \[ \boxed{x=4}_{\ \in \text{D}} \quad \lor \quad x=-8_{\ \notin \text{D}} \]
Wyznaczamy dziedzinę: \[ \begin{split} 8-x^{3}\gt 0\ &\land\ 2-x\gt 0\ &\land\ \log(2-x)\ne 0\\[6pt] x\lt 2\ &\land\ x\lt 2\ &\land\ x\ne 1 \end{split} \] Zatem: \[ x\lt 2\quad \land\quad x\ne 1 \]
Stosujemy wzór na zamianę podstaw logarytmów: \[ \log_{\,2-x}(8-x^3)=3\\[6pt] (2-x)^3=8-x^3\\[6pt] 8-12x+6x^2-x^3=8-x^3\\[6pt] 6x^2-12x=0\\[6pt] x(x-2)=0 \] Zatem: \[ \boxed{x=0}_{\ \in \text{D}} \quad \lor \quad x=2_{\ \notin \text{D}} \]
Wyznaczamy dziedzinę: \[ \begin{split} 3x-10\gt 0\ &\land\ 5x-12\gt 0\ &\land\ x\gt 0\ &\land\ \ln x\ne 0\\[6pt] x\gt \tfrac{10}{3}\ &\land\ x\gt \tfrac{12}{5}\ &\land\ x\gt 0\ &\land\ x\ne 1 \end{split} \] Zatem: \[ x\gt \tfrac{10}{3} \]
Odejmujemy logarytmy i porównujemy liczby logarytmowane: \[ \ln\!\left(\frac{3x-10}{5x-12}\right)=-\ln x\\[6pt] \frac{3x-10}{5x-12}=\frac{1}{x}\\[6pt] x(3x-10)=5x-12\\[6pt] 3x^{2}-15x+12=0\\[6pt] \Delta=(-15)^2-4\cdot 3\cdot 12=225-144=81 \] Zatem: \[ x=1 \quad \lor \quad \boxed{x=4}_{\ \in \text{D}} \] ( \(x=1_{\ \notin \text{D}}\) ).