Bài 3 (Trang 84 SGK Toán Giải Tích 12):

Giải các phương trình lôgarit :

a) ${\log }_3\left(5x + 3\right) = {\log }_3\left(7x + 5\right)$;

b) $\log \left(x-1\right)- \log \left(2x-11\right) = \log 2$;

c) ${\log }_2\left(x-5\right) + {\log }_2\left(x+2\right) = 3$;

d) $\log \left(x^2-6x+7\right) = \log \left(x-3\right)$.

Hướng dẫn giải:

a) Điều kiện $\left\{\begin{array}{l}5\mathrm{x}+3 >0\\ 7\mathrm{x}+5 >0\end{array}\right.\iff \mathrm{x}>-\frac{3}{5}$

${\log }_3\left(5\mathrm{x}+3\right) = {\log }_3\left(7\mathrm{x}+5\right)\iff 5\mathrm{x}+3 = 7\mathrm{x}+5\iff \mathrm{x} = -1 \left(\mathrm{loại}\right)$

Vậy $S = \cancel{O}$

b) Điều kiện $\left\{\begin{array}{l}\mathrm{x}-1>0\\ 2\mathrm{x}-11>0\end{array}\right.\iff \mathrm{x}>\frac{11}{2}$

$\log \left(x-1\right)- \log \left(2x-11\right) = \log 2$

$\iff \log \frac{\mathrm{x}-1}{2\mathrm{x}-11} = \log 2\iff \frac{\mathrm{x}-1}{2\mathrm{x}-11} = 2\iff \mathrm{x}-1 = 4\mathrm{x}-22\iff \mathrm{x} = 7 \left(\mathrm{nhận}\right)$

Vậy $S = \left\{\left.7\right\}\right.$

c) Điều kiện: $x>5$

Ta có:   $$\begin{gathered} {\log }_2\left(\mathrm{x}-5\right)+{\log }_2\left(\mathrm{x}+2\right) = 3\iff {\log }_2\left(\mathrm{x}-5\right)\left(\mathrm{x}+2\right) = {\log }_{2}8 \iff \left(\mathrm{x}-5\right)\left(\mathrm{x}+2\right) = 8\iff {\mathrm{x}}^2 - 3\mathrm{x}- 18=0 \\ \iff [\begin{array}{rcl}\mathrm{x}& =& 6\\ \mathrm{x}& =& -3 \left(\mathrm{Loại}\right)\end{array} \end{gathered}$$

Vậy S = {6}.

d) Ta có: $\log \left(x^2-6x+7\right) = \log (x-3)$

$\iff \left\{\begin{array}{l}\mathrm{x}-3>0\\ {\mathrm{x}}^2-6\mathrm{x}+7 = \mathrm{x}-3\end{array}\iff \left\{\begin{array}{l}\mathrm{x}>3\\ {\mathrm{x}}^2-7\mathrm{x}+10 = 0\end{array}\right.\right.\iff \mathrm{x} = 5$

Vậy S = {5}.