LG a
\(\eqalign{
{\log _2}\left( {3 - x} \right) + {\log _2}\left( {1 - x} \right) = 3; \cr } \)
Lời giải chi tiết:
Điều kiện:
\(\left\{ \begin{array}{l}
3 - x > 0\\
1 - x > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x < 3\\
x < 1
\end{array} \right. \Leftrightarrow x < 1\)
\(\eqalign{
& {\log _2}\left( {3 - x} \right) + {\log _2}\left( {1 - x} \right) = 3 \cr&\Leftrightarrow {\log _2}[\left( {3 - x} \right)\left( {1 - x} \right)] = 3 \cr
& \Leftrightarrow \left( {3 - x} \right)\left( {1 - x} \right) = 8 \cr& \Leftrightarrow {x^2} - 4x - 5 = 0\left[ \matrix{
x = - 1(TM) \hfill \cr
x = 5\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ { - 1} \right\}\)
LG b
\(\eqalign{
{\log _2}\left( {9 - {2^x}} \right) = {10^{\log \left( {3 - x} \right)}} \cr} \)
Lời giải chi tiết:
Điều kiện:
\(\begin{array}{l}
\left\{ \begin{array}{l}
3 - x > 0\\
9 - {2^x} > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x < 3\\
{2^x} < 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x < 3\\
x < {\log _2}9
\end{array} \right. \Leftrightarrow x < 3
\end{array}\)
\(\eqalign{
& {\log _2}\left( {9 - {2^x}} \right) = {10^{\log \left( {3 - x} \right)}} \cr
&\Leftrightarrow {\log _2}\left( {9 - {2^x}} \right) = 3 - x \cr
&\Leftrightarrow 9 - {2^x} = {2^{3 - x}} \cr
& \Leftrightarrow 9 - {2^x} = {8 \over {{2^x}}}\cr
& \Leftrightarrow {9.2^x} - {\left( {{2^x}} \right)^2} = 8 \cr&\Leftrightarrow {\left( {{2^x}} \right)^2} - {9.2^x} + 8 = 0 \cr
&\Leftrightarrow \left[ \matrix{
{2^x} = 1 \hfill \cr
{2^x} = 8 \hfill \cr} \right. \cr
&\Leftrightarrow \left[ \matrix{
x = 0(TM) \hfill \cr
x = 3\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
LG c
\(\eqalign{
{7^{\log x}} - {5^{\log x + 1}} = {3.5^{\log x - 1}} - 13.{7^{\log x - 1}} \cr} \)
Lời giải chi tiết:
Điều kiện: \(x > 0\)
\(\begin{array}{l}
PT \Leftrightarrow {7.7^{\log x - 1}} - {5^2}{.5^{\log x - 1}} = {3.5^{\log x - 1}} - {13.7^{\log x - 1}}\\
\Leftrightarrow {7.7^{\log x - 1}} + {13.7^{\log x - 1}} = {3.5^{\log x - 1}} + {25.5^{\log x - 1}}\\
\Leftrightarrow {20.7^{\log x - 1}} = {28.5^{\log x - 1}}\\
\Leftrightarrow \frac{{{7^{\log x - 1}}}}{{{5^{\log x - 1}}}} = \frac{{28}}{{20}}\\
\Leftrightarrow {\left( {\frac{7}{5}} \right)^{\log x - 1}} = \frac{7}{5}\\
\Leftrightarrow \log x - 1 = 1\\
\Leftrightarrow \log x = 2\\
\Leftrightarrow x = {10^2} = 100\left( {TM} \right)
\end{array}\)
Vậy \(S = \left\{ {100} \right\}\)
LG d
\(\eqalign{
{6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr} \)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& {6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr
& \Leftrightarrow {6^x} + {6.6^x} = {2^x} + {2.2^x} + {2^2}{.2^x}\cr&\Leftrightarrow {6^x}\left( {1 + 6} \right) = {2^x}\left( {1 + 2 + {2^2}} \right) \cr
& \Leftrightarrow {6^x}.7 = {2^x}.7\cr& \Leftrightarrow \frac{{{6^x}}}{{{2^x}}} = \frac{7}{7}\cr&\Leftrightarrow {3^x} = 1 \cr
& \Leftrightarrow x = 0 \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
