Giải thích các bước giải:
ĐK: x>-1
Ta có:
\[\begin{array}{l}
{\log _2}\left( {x + 1} \right) = {\log _4}\left( {x + 2} \right)\\
\Leftrightarrow {\log _2}\left( {x + 1} \right) = {\log _{{2^2}}}\left( {x + 2} \right)\\
\Leftrightarrow {\log _2}\left( {x + 1} \right) = \frac{1}{2}{\log _2}\left( {x + 2} \right)\\
\Leftrightarrow {\log _2}\left( {x + 1} \right) = {\log _2}\sqrt {x + 2} \\
\Leftrightarrow x + 1 = \sqrt {x + 2} \\
\Leftrightarrow {x^2} + 2x + 1 = x + 2\\
\Leftrightarrow {x^2} + x - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 1 + \sqrt 5 }}{2}\\
x = \frac{{ - 1 - \sqrt 5 }}{2}
\end{array} \right.
\end{array}\]
