Bài tập 75 trang 127 SGK Toán 12 NC

Bài tập 75 trang 127 SGK Toán 12 NC

a) Điều kiện: x > 0

Ta có: \({\log _3}\left( {{3^x} - 1} \right).{\log _3}\left( {{3^{x + 1}} - 3} \right) = 12\)

\(\begin{array}{*{20}{l}}
{ \Leftrightarrow {{\log }_3}\left( {{3^x} - 1} \right).{{\log }_3}3\left( {{3^x} - 1} \right) = 12}\\
{ \Leftrightarrow {{\log }_3}\left( {{3^x} - 1} \right)\left[ {1 + {{\log }_3}\left( {{3^x} - 1} \right)} \right] = 12}\\
{ \Leftrightarrow \log _3^2\left( {{3^x} - 1} \right) + {{\log }_3}\left( {{3^x} - 1} \right) - 12 = 0}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_3}({3^x} - 1) =  - 4}\\
{{{\log }_3}({3^x} - 1) = 3}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = {{\log }_3}\frac{{82}}{{81}}}\\
{x = {{\log }_3}28}
\end{array}} \right.}
\end{array}\)

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

b) Điều kiện: x > 1; x ≠ 2

Ta có: 

\({\log _{x - 1}}4 = \frac{1}{{{{\log }_4}\left( {x - 1} \right)}} = \frac{2}{{{{\log }_2}\left( {x - 1} \right)}}\)

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

Ta có phương trình:

\(\begin{array}{*{20}{l}}
{\frac{2}{t} = 1 + t \Leftrightarrow {t^2} + t - 2 = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{t = 1}\\
{t =  - 2}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_2}(x - 1) = 1}\\
{{{\log }_2}(x - 1) =  - 2}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 3}\\
{x = \frac{5}{4}}
\end{array}} \right.
\end{array}
\end{array}\)

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

c) Điều kiện:

\(\begin{array}{l}
{\log _2}( - x) \ge 0\\
 \Leftrightarrow  - x \ge 1 \Leftrightarrow x \le  - 1
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\sqrt {{x^2}} \\
 \Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\left( { - x} \right)
\end{array}\\
{ \Leftrightarrow 5\sqrt t  = t,\left( {t = {{\log }_2}\left( { - x} \right) \ge 0} \right)}\\
\begin{array}{l}
 \Leftrightarrow 25t = {t^2} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{t = 0}\\
{t = 25}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_2}( - x) = 0}\\
{{{\log }_2}( - x) = 25}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - 1}\\
{x =  - {2^{25}}}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(S = \{  - 1; - {2^{25}}\} \)

d) Điều kiện: x > 0

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

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{3^{\frac{1}{2} + {{\log }_4}x}} + {3^{{{\log }_4}x - \frac{1}{2}}} = \sqrt x \\
 \Leftrightarrow \left( {\sqrt 3  + \frac{1}{{\sqrt 3 }}} \right){3^{{{\log }_4}x}} = {2^{{{\log }_4}x}}
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \frac{4}{{\sqrt 3 }} = {\left( {\frac{2}{3}} \right)^{{{\log }_4}x}}\\
 \Leftrightarrow {\log _4}x = {\log _{\frac{2}{3}}}\frac{4}{{\sqrt 3 }}
\end{array}\\
{ \Leftrightarrow x = {4^{{{\log }_{\frac{2}{3}}}\frac{4}{{\sqrt 3 }}}}}
\end{array}\)

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

-- Mod Toán 12