|
|
Điều kiện : $x > 0$
Ta có : ${\log _2}x + {\log _4}x + {\log _8}x = 11$
$\Leftrightarrow $ ${\log _2}x + {\log _{{2^2}}}x + {\log _{{2^3}}}x = 11$
$\Leftrightarrow $ ${\log _2}x + \frac{1}{2}{\log _2}x + \frac{1}{3}{\log _2}x = 11$
$\Leftrightarrow $ $\frac{{11}}{6}{\log _2}x = 11 \Leftrightarrow {\log _2}x = 6 \Leftrightarrow
x = {2^6} = 64$ (nhận)
|