Bài 75 trang 127 SGK giải tích 12 nâng cao

Đề bài

Bài 75 

\(\eqalign{
& a)\,{\log _3}\left( {{3^x} - 1} \right).{\log _3}\left( {{3^{x + 1}} - 3} \right) = 12; \cr 
& c)\,5\sqrt {{{\log }_2}\left( { - x} \right)} = {\log _2}\sqrt {{x^2}} ; \cr} \)   

\(\eqalign{
& b)\,{\log _{x - 1}}4 = 1 + {\log _2}\left( {x - 1} \right); \cr
& d)\,{3^{{{\log }_4} + {1 \over 2}}} + \,{3^{{{\log }_4} - {1 \over 2}}} = \sqrt x . \cr} \)                               

Hướng dẫn giải

a) Điều kiện: \(x > 0\)

Ta có: \(lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}\left( {{3^{x + 1}} - 3} \right) = 12\) 

\(\eqalign{
& \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}3\left( {{3^x} - 1} \right) = 12 \cr
& \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right)\left[ {1 + lo{g_3}\left( {{3^x} - 1} \right)} \right] = 12 \cr} \)

 \( \Leftrightarrow \log _3^2\left( {{3^x} - 1} \right) + lo{g_3}\left( {{3^x} - 1} \right) - 12 = 0\) 

\(\eqalign{
& \Leftrightarrow \left[ \matrix{
lo{g_3}\left( {{3^x} - 1} \right) = - 4 \hfill \cr
lo{g_3}\left( {{3^x} - 1} \right) = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{3^x} - 1 = {1 \over {81}} \hfill \cr
{3^x} - 1 = {3^3} = 27 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{3^x} = {{82} \over {81}} \hfill \cr
{3^x} = 28 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\log _3}{{82} \over {81}} \hfill \cr
x = {\log _3}28 \hfill \cr} \right. \cr} \)

Vậy \(S = \left\{ {{{\log }_3}28;{{\log }_3}82 - 4} \right\}\)

b) Điều kiện: \(x > 1\); \(x \ne 2\)

Ta có: \({\log _{x - 1}}4 = {1 \over {{{\log }_4}\left( {x - 1} \right)}} = {2 \over {{{\log }_2}\left( {x - 1} \right)}}\). Đặt \(t = {\log _2}\left( {x - 1} \right)\)

Ta có phương trình:

\(\eqalign{
& {2 \over t} = 1 + t \Leftrightarrow {t^2} + t - 2 = 0 \cr
& \Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = - 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _2}\left( {x - 1} \right) = 1 \hfill \cr
{\log _2}\left( {x - 1} \right) = - 2 \hfill \cr} \right.\left[ \matrix{
x = 3 \hfill \cr
x = {5 \over 4} \hfill \cr} \right. \cr} \)

Vậy \(S = \left\{ {3;{5 \over 4}} \right\}\)

c) Điều kiện: \({\log _2}\left( { - x} \right) \ge 0 \Leftrightarrow  - x \ge 1 \Leftrightarrow x \le  - 1\)

    \(5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\sqrt {{x^2}}  \Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\left( { - x} \right)\)    

 \( \Leftrightarrow 5\sqrt t  = t\) với \(t = {\log _2}\left( { - x} \right) \ge 0\) 

\(\eqalign{
& \Leftrightarrow 25t = {t^2} \Leftrightarrow \left[ \matrix{
t = 0 \hfill \cr
t = 25 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _2}\left( { - x} \right) = 0 \hfill \cr
lo{g_2}\left( { - x} \right) = 25 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = - 1 \hfill \cr
x = - {2^{25}} \hfill \cr} \right. \cr} \)

Vậy \(S = \left\{ { - 1; - {2^{25}}} \right\}\)

d) Điều kiện: \(x > 0\)

Ta có: \(\sqrt x  = \sqrt {{4^{{{\log }_4}x}}}  = {2^{{{\log }_4}x}}\)

Do đó \({3^{{1 \over 2} + {{\log }_4}x}} + {3^{{{\log }_4}x - {1 \over 2}}} = \sqrt x  \Leftrightarrow \left( {\sqrt 3  + {1 \over {\sqrt 3 }}} \right){3^{{{\log }_4}x}} = {2^{{{\log }_4}x}}\) 

\(\eqalign{
& \Leftrightarrow {4 \over {\sqrt 3 }} = {\left( {{2 \over 3}} \right)^{{{\log }_4}x}} \Leftrightarrow {\log _4}x = {\log _{{2 \over 3}}}{4 \over {\sqrt 3 }} \cr
& \Leftrightarrow x = {4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}} \cr} \)

Vậy \(S = \left\{ {{4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}}} \right\}\)