Đáp án:
\[x = 2\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x - 1 > 0\\
x + 3 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
x > - 3
\end{array} \right. \Leftrightarrow x > 1\)
Ta có:
\(\begin{array}{l}
{\log _a}b + {\log _a}c = {\log _a}\left( {bc} \right)\\
{\log _a}b - {\log _a}c = {\log _a}\frac{b}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a > 0,a \ne 1,\,\,b > 0,c > 0} \right)\\
{\log _2}\left( {x - 1} \right) + {\log _2}\left( {x + 3} \right) = {\log _2}10 - 1\\
\Leftrightarrow {\log _2}\left( {x - 1} \right) + {\log _2}\left( {x + 3} \right) = {\log _2}10 - {\log _2}2\\
\Leftrightarrow {\log _2}\left[ {\left( {x - 1} \right)\left( {x + 3} \right)} \right] = {\log _2}\frac{{10}}{2}\\
\Leftrightarrow {\log _2}\left[ {\left( {x - 1} \right)\left( {x + 3} \right)} \right] = {\log _2}5\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 3} \right) = 5\\
\Leftrightarrow {x^2} + 2x - 3 = 5\\
\Leftrightarrow {x^2} + 2x - 8 = 0\\
\Leftrightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4 = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 4\\
x = 2
\end{array} \right.\\
x > 1 \Rightarrow x = 2
\end{array}\)
Vậy \(x = 2\)
