Bài 75 trang 127 SGK giải tích 12 nâng cao

LG a

\(\eqalign{
{\log _3}\left( {{3^x} - 1} \right).{\log _3}\left( {{3^{x + 1}} - 3} \right) = 12; \cr} \)   

Lời giải chi tiết:

Điều kiện:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{3^x} - 1 > 0\\
{3^{x + 1}} - 3 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{3^x} - 1 > 0\\
{3.3^x} - 3 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{3^x} - 1 > 0\\
3\left( {{3^x} - 1} \right) > 0
\end{array} \right. \Leftrightarrow {3^x} - 1 > 0\\
\Leftrightarrow {3^x} > 1 \Leftrightarrow x > 0
\end{array}\)

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

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

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

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

Vậy \(S = \left\{ {{{\log }_3}28;{\log _3}{{82} \over {81}} } \right\}\)

LG b

\(\eqalign{
{\log _{x - 1}}4 = 1 + {\log _2}\left( {x - 1} \right); \cr} \)    

Lời giải chi tiết:

Điều kiện: \(0 < x - 1 \ne 1 \Leftrightarrow 1 < x \ne 2\)

Ta có: \({\log _{x - 1}}4 = {1 \over {{{\log }_4}\left( {x - 1} \right)}} \)

\( = \frac{1}{{{{\log }_{{2^2}}}\left( {x - 1} \right)}} = \frac{1}{{\frac{1}{2}{{\log }_2}\left( {x - 1} \right)}}\)

\(= {2 \over {{{\log }_2}\left( {x - 1} \right)}}\).

Đặt \(t = {\log _2}\left( {x - 1} \right)\)

Ta có phương trình:

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

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

LG c

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

Lời giải chi tiết:

Điều kiện:

\(\left\{ \begin{array}{l}
- x > 0\\
{\log _2}\left( { - x} \right) \ge 0\\
\sqrt {{x^2}} > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x < 0\\
- x \ge {2^0} = 1\\
x \ne 0
\end{array} \right. \) \(\Leftrightarrow \left\{ \begin{array}{l}
x < 0\\
x \le - 1
\end{array} \right. \Leftrightarrow x \le - 1\)

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

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

\(\Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\left( { - x} \right)\) (vì \(x \le  - 1 \Rightarrow \left| x \right| =  - x\))

Đặt \(t = {\log _2}\left( { - x} \right) \ge 0\) ta được:

\(\eqalign{
& 5\sqrt t  = t \Leftrightarrow 25t = {t^2} \cr &\Leftrightarrow \left[ \matrix{
t = 0 \hfill \cr 
t = 25 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _2}\left( { - x} \right) = 0 \hfill \cr 
lo{g_2}\left( { - x} \right) = 25 \hfill \cr} \right. \cr 
& \Leftrightarrow \left[ \matrix{
x = - 1 \hfill \cr 
x = - {2^{25}} \hfill \cr} \right. \cr} \)

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

LG d

\(\eqalign{
{3^{{{\log }_4} x+ {1 \over 2}}} + \,{3^{{{\log }_4} x- {1 \over 2}}} = \sqrt x . \cr} \)    

Lời giải chi tiết:

Điều kiện: \(x > 0\)

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

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

\(\Leftrightarrow {3^{\frac{1}{2}}}{.3^{{{\log }_4}x}} + {3^{{{\log }_4}x}}{.3^{ - \frac{1}{2}}} = {2^{{{\log }_4}x}}\)

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

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

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

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