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Bài 4. Giải các phương trình lôgarit:
a) \({1 \over 2}\log \left( {{x^2} + x – 5} \right) = \log 5{\rm{x}} + \log {1 \over {5{\rm{x}}}}\)
b) \({1 \over 2}\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = \log 8{\rm{x}} – \log 4{\rm{x}}\)
c) \({\log _{\sqrt 2 }}x + 4{\log _{4{\rm{x}}}}x + {\log _8}x = 13\)
a) \({1 \over 2}\log \left( {{x^2} + x – 5} \right) = \log 5{\rm{x}} + \log {1 \over {5{\rm{x}}}}\)
\(\Leftrightarrow \left\{ \matrix{
5{\rm{x}} > 0 \hfill \cr
{1 \over 2}\log \left( {{x^2} + x – 5} \right) = \log 5{\rm{x}} – \log 5{\rm{x}}\hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
x > 0 \hfill \cr
{1 \over 2}\log \left( {{x^2} + x – 5} \right) = 0 \hfill \cr} \right.\)
\(\Leftrightarrow\left\{ \matrix{
x > 0 \hfill \cr
\log \left( {{x^2} + x – 5} \right) = 0 \hfill \cr} \right.\)
\(\Leftrightarrow\left\{ \matrix{
x > 0 \hfill \cr
{x^2} + x – 5 = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x > 0 \hfill \cr
{x^2} + x – 6 = 0 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
x > 0 \hfill \cr
x = – 3;x = 2 \hfill \cr} \right. \Leftrightarrow x = 2\)
Vậy nghiệm của phương trình là \(x = 2\)
b) \({1 \over 2}\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = \log 8{\rm{x}} – \log 4{\rm{x}}\)
\(\Leftrightarrow\left\{ \matrix{
4{\rm{x > 0}} \hfill \cr
{{\rm{x}}^2} – 4{\rm{x}} – 1 > 0 \hfill \cr
{1 \over 2}\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = \log {{8{\rm{x}}} \over {4{\rm{x}}}} \hfill \cr} \right.\)
\(\Leftrightarrow\left\{ \matrix{
x > 0 \hfill \cr
{{\rm{x}}^2} – 4{\rm{x}} – 1 > 0 \hfill \cr
{1 \over 2}\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = \log 2 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
x > 0 \hfill \cr
\left[ \matrix{
x > 2 + \sqrt 5 \hfill \cr
x < 2 – \sqrt 5 \hfill \cr} \right. \hfill \cr
\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = 2\log 2 \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{
x > 2 + \sqrt 5 \hfill \cr
\log \left( {{x^2} – 4{\rm{x}} – 1} \right) = \log {2^2} = \log 4 \hfill \cr} \right.\)
\(\Leftrightarrow\left\{ \matrix{
x > 2 + \sqrt 5 \hfill \cr
{x^2} – 4{\rm{x}} – 1 = 4 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
x > 2 + \sqrt 5 \hfill \cr
{x^2} – 4{\rm{x}} – 5 = 0 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
x > 2 + \sqrt 5 \hfill \cr
x = – 1;x = 5 \hfill \cr} \right. \Leftrightarrow x = 5\)
Vậy nghiệm của phương trình là \(x = 5\)
c) \({\log _{\sqrt 2 }}x + 4{\log _{4}}x + {\log _8}x = 13\)
\(\Leftrightarrow {\log _{{2^{{1 \over 2}}}}}x + 4{\log _{{2^2}}}x + {\log _{{2^3}}}x = 13\)
\(\Leftrightarrow 2{\log _2}x + 2{\log _2}x + {1 \over 3}{\log _2}x = 13\)
\(\Leftrightarrow {{13} \over 3}{\log _2}x = 13 \Leftrightarrow {\log _2}x = 3 \Leftrightarrow x = {2^3} = 8\)
Vậy phương trình có nghiệm là \(x = 8\)