a) Biết $\log_712 = a, \log_{12}24 = b$. Tính $\log_{54}168$ theo $a$ và $b$.
b) Biết $\log_615 = a, \log_{12}18 = b$. Tính $ \log_{25}24$ theo  $a$ và $b$.
a) Ta có $\log_{54}168=\frac{\log_{7}168}{\log_{7}54} =\frac{\log_{7}(3.7.2^3)}{\log_{7}(2.3^3)}=\frac{\log_{7}3+1+3\log_{7}2}{\log_{7}2+3\log_{7}3}  $
(Ta phi tính $\log_{7}2$ và $\log_{7}3$ theo $a=\log_{7}12 $ và $b=\log_{12}24$)
T gi thiết $\left\{ \begin{array}{l} a=\log_{7}12 \\ b=\log_{12}24  \end{array} \right.\Rightarrow \left\{ \begin{array}{l} a=\log_{7}(2^2.3)=2\log_{7}2+\log_{7}3  \\ ab=\log_{7}12.\log_{12}24=\log_{7}24    \end{array} \right.  $
Hay $\left\{ \begin{array}{l} a=2\log_{7}2+\log_{7}3  \\ ab=\log_{7}(2^3.3)=3\log_{7}2+\log_{7}3    \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \log_{7}2=ab-a \\ \log_{7}3=3a-2ab  \end{array} \right.  $
Sau cùng ta được $     \log_{54}168=  \frac{(3a - 2ab) + 1 + 3 (ab - a)}{(ab - a) + 3 (3a- 2ab)} $
                                  $\Leftrightarrow \log_{54}168=\frac{ab+1}{a(8-5b)}$


b) Ta có $\log_{25}24=\log_{5^2}(2^3.3)=\frac{1}{2}(3\log_{5}2+\log_{5}3)$

(Ta phi tính $\log_{5}2 $ và $\log_{5}3 $ theo $a=\log_{6}15$ và $\log_{12}18)$.
Ta có $\left\{ \begin{array}{l} a=\log_{6}15=\frac{\log_{5}15 }{\log_{5}6 }=\frac{1+\log_{5}3 }{\log_{5}2+\log_{5}3}                                                              (1)\\b=\log_{12}18=\frac{\log_{5}18 }{\log_{5}12 }= \frac{\log_{5}(2.3^2)}{\log_{5}(2^2.3)}=\frac{\log_{5}2+2\log_{5}3} {2\log_{5}2+\log_{5}3 }              (2)\end{array} \right.$  
T $(1)$ và $(2)$ ta có $\left\{ \begin{array}{l} a=\frac{1+y}{x+y} \\ b=\frac{x+2y}{2x+y}  \end{array} \right.$ 
với  $\left\{ \begin{array}{l} x=\log_{5}2 \\ y=\log_{5}3  \end{array} \right.$ hay $\left\{ \begin{array}{l} ax+(a-1)y = 1\\ (2b-1)x+(b-2)y=0 \end{array} \right.$
$D =  \left| \begin{array}{l}a              a-1\\2b - 1      b-2\end{array} \right| $    =$-a-ab+2b-1$;
$D_{x} = \left| \begin{array}{l}1         a-1\\0         b-2\end{array} \right| $=  $b-2$;  
$D_{y}=\left| \begin{array}{l}a                     1\\2b-1      0\end{array} \right|$=  $1-2b$
 
Vy $\left\{ \begin{array}{l} x=\frac{D_{x}}{D}=\frac{b-2}{-a-ab+2b-1}=\frac{2-b}{a+ab-2b+1}   \\ y=\frac{D_{y}}{D}=\frac{1-2b}{-a-ab+2b-1}=\frac{2b-1}{a+ab-2b+1}    \end{array} \right.  $   
Lúc đó, $\log_{25}24=\frac{1}{2}(3x+y)=\frac{1}{2}(\frac{6-3b+2b-1}{a+ab-2b+1})    $
Hay $\log_{25}24=\frac{1}{2}(\frac{5-b}{a+ab-2b+1})   $
có cách nào ngắn hơn k hả chị? –  bautroixanh_0801 16-07-12 01:52 PM

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