Tính :
$\begin{array}{l}
1)   A = {\log _3}2.{\log _4}3.{\log _5}4...{\log _{15}}14.{\log _{16}}15\\

\end{array}$
$2)   {\log _6}16$   
Cho biết  ${\log _{12}}27 = X$
$1)A = {\log _{16}}15.{\log _{15}}14...{\log _5}4.{\log _4}3.{\log _3}2 = {\log _{16}}2 = \frac{1}{4}$
2)Chọn $2$ là cơ số ta có:
${\log _6}16 = \frac{{{{\log }_2}16}}{{{{\log }_2}6}} = \frac{4}{{1 + {{\log }_2}3}}$

Ta có:
$\begin{array}{l}
X = {\log _{12}}27 = \frac{{{{\log }_2}27}}{{{{\log }_2}12}} = \frac{{3{{\log }_2}3}}{{2 + {{\log }_2}3}}\\
 \Rightarrow {\log _2}3 = \frac{{2X}}{{3 - X}} \Rightarrow {\log _6}16 = \frac{{4(3 - X)}}{{3 + X}}
\end{array}$

Thẻ

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