Bài tập 70 trang 125 SGK Toán 12 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{3^{4x}} = {4^{3x}}\\
 \Leftrightarrow {4^x}{\log _3}3 = {3^x}{\log _3}4\\
 \Leftrightarrow \frac{{{4^x}}}{{{3^x}}} = {\log _3}4
\end{array}\\
\begin{array}{l}
 \Leftrightarrow {\left( {\frac{4}{3}} \right)^x} = {\log _3}4\\
 \Leftrightarrow x = {\log _{\frac{4}{3}}}({\log _3}4)
\end{array}
\end{array}\)

Vậy \(S = \left\{ {{{\log }_{\frac{4}{3}}}({{\log }_3}4)} \right\}\)

b) Điều kiện x > 0

\(\begin{array}{*{20}{l}}
{{3^{2 - {{\log }_3}x}} = 81x \Leftrightarrow \frac{{{3^2}}}{{{3^{lo{g_3}x}}}} = 81x}\\
\begin{array}{l}
 \Leftrightarrow \frac{9}{x} = 81x \Leftrightarrow {x^2} = \frac{1}{8}\\
 \Leftrightarrow x = \frac{1}{3}\left( {x > 0} \right)
\end{array}
\end{array}\)

Vậy S = {1/3}

c) Lấy logarit cơ số 3 hai vế ta được:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
x{\log _3}3 + \frac{x}{{x + 1}}{\log _3}8 = 2 + 2.{\log _3}2\\
 \Leftrightarrow x + \frac{{3x}}{{x + 1}}{\log _3}2 = 2 + 2.{\log _3}2
\end{array}\\
\begin{array}{l}
 \Leftrightarrow {x^2} + x + 3({\log _3}2)x\\
 = 2x + 2 + 2(x + 1)({\log _3}2)
\end{array}\\
{ \Leftrightarrow {x^2} + ({{\log }_3}2 - 1)x - 2.{{\log }_3}2 - 2 = 0}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 2}\\
{x =  - 1 - {{\log }_3}2}
\end{array}} \right.}
\end{array}\)

Vậy \(S = \left\{ {2; - 1 - {{\log }_3}2} \right\}\)

d) Điều kiện x > 0

Lấy logarit cơ số x hai vế ta được:

\(\begin{array}{*{20}{l}}
{6 + ( - {{\log }_x}5).{{\log }_x}5 =  - 5{{\log }_x}5}\\
{ \Leftrightarrow \log _x^25 - 5{{\log }_x}5 - 6 = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_x}5 =  - 1}\\
{{{\log }_x}5 = 6}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{5 = {x^{ - 1}}}\\
{5 = {x^6}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \frac{1}{5}}\\
{x = \sqrt[6]{5}}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(S = \left\{ {\frac{1}{5};\sqrt[6]{5}} \right\}\)

-- Mod Toán 12 HỌC247