Bài 94 trang 131 SGK giải tích 12 nâng cao

Bài 94

\(\eqalign{
& a)\,{\log _3}\left( {\log _{0,5}^2x – 3{{\log }_{0,5}}x + 5} \right) = 2\,; \cr
& c)\,1 – {1 \over 2}\log \left( {2x – 1} \right) = {1 \over 2}\log \left( {x – 9} \right)\,; \cr} \)

\(\eqalign{
& b)\,{\log _2}\left( {{{4.3}^x} – 6} \right) – {\log _2}\left( {{9^x} – 6} \right) = 1\,; \cr
& d)\,{1 \over 6}{\log _2}\left( {x – 2} \right) – {1 \over 3} = {\log _{{1 \over 8}}}\sqrt {3x – 5}. \cr} \)

Giải

\(\eqalign{
& a)\,\,{\log _3}\left( {\log _{0,5}^2x – 3{{\log }_{0,5}}x + 5} \right) = 2 \Leftrightarrow \log _{0,5}^2x – 3{\log _{0,5}}x + 5 = 9 \cr
& \Leftrightarrow \log _{0,5}^2x – 3{\log _{0,5}x} – 4 = 0 \Leftrightarrow \left[ \matrix{
{\log _{0,5}x} = – 1 \hfill \cr
{\log _{0,5}x} = 4 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\left( {0,5} \right)^{ – 1}} = 2 \hfill \cr
x = {\left( {0,5} \right)^4} = {1 \over {16}} \hfill \cr} \right. \cr} \)

Vậy \(S = \left\{ {2;{1 \over {16}}} \right\}\)

b) Ta có: \({\log _2}\left( {{{4.3}^x} – 6} \right) – {\log _2}\left( {{9^x} – 6} \right) = 1 \Leftrightarrow {\log _2}\left( {{{4.3}^x} – 6} \right) = {\log _2}2\left( {{9^x} – 6} \right)\)

\( \Leftrightarrow \left\{ \matrix{
{9^x} – 6 > 0 \hfill \cr
{4.3^x} – 6 = 2\left( {{9^x} – 6} \right) \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
t > \sqrt 6 \hfill \cr
2{t^2} – 4t – 6 = 0 \hfill \cr} \right.\) (với \(t = {3^x}\))

\( \Leftrightarrow t = 3 \Leftrightarrow {3^x} = 3 \Leftrightarrow x = 1\)

Vậy \(S = \left\{ 1 \right\}\)
c) Điều kiện: \(x >9\)

\(\eqalign{
& 1 – {1 \over 2}\log \left( {2x – 1} \right) = {1 \over 2}\log \left( {x – 9} \right) \Leftrightarrow 2 = \log \left( {2x – 1} \right) + \log \left( {x – 9} \right) \cr
& \Leftrightarrow \log \left( {2x – 1} \right)\left( {x – 9} \right) = 2 \Leftrightarrow \left( {2x – 1} \right)\left( {x – 9} \right) = 100 \cr
& \Leftrightarrow 2{x^2} – 19x – 91 = 0 \Leftrightarrow \left[ \matrix{
x = 13 \hfill \cr
x = – 3,5\,\,\left( \text {loại} \right) \hfill \cr} \right. \cr} \)

Vậy \(x=13\)

d) Điều kiện: \(x > 2\)

Ta có: \({\log _{{1 \over 8}}}\sqrt {3x – 5}  = {\log _{{2^{ – 3}}}}{\left( {3x – 5} \right)^{{1 \over 2}}} =  – {1 \over 6}{\log _2}\left( {3x – 5} \right)\)
Phương trình đã có trở thành:

\(\eqalign{
& {1 \over 6}{\log _2}\left( {x – 2} \right) + {1 \over 6}{\log _2}\left( {3x – 5} \right) = {1 \over 3} \cr
& \Leftrightarrow {\log _2}\left( {x – 2} \right)\left( {3x – 5} \right) = 2 \cr
& \Leftrightarrow \left( {x – 2} \right)\left( {3x – 5} \right) = 4 \cr
& \Leftrightarrow x = 3\,\,\text{ hoặc }\,\,x = {2 \over 3}. \cr} \)

Với điều kiện \(x > 2\) ta chỉ nhận nghiệm \(x = 3\).
Vậy \(S = \left\{ 3 \right\}\)