2 replies to this topic

[Toanhochoctoan]

Cho $\log_7{12}=x, \log_{12}24=y$ và $\log_{54}{168}=\frac{axy+1}{bxy+cx}$, trong đó $a,b,c$ nguyên. Tính $S=a+2b+3c$

[trambau]

Cho $\log_7{12}=x, \log_{12}24=y$ và $\log_{54}{168}=\frac{axy+1}{bxy+cx}$, trong đó $a,b,c$ nguyên. Tính $S=a+2b+3c$

$\left\{\begin{matrix} \log_712=x & & \\ \log_{12}24=y & & \end{matrix}\right. \Rightarrow xy=\log_712.\log_{12}24=\log_724 \Rightarrow \log_{54}168=\frac{\log_7168}{\log_754}=\frac{\log_724+\log_77}{\log_754}=\frac{xy+1}{\log_754}\Rightarrow a=1$

Suy ra $bxy+cx=\log_754\Leftrightarrow b.\log_754+c.\log_712=\log_754\Leftrightarrow \log_7(24^b.12^c)=\log_754\Leftrightarrow c=\log_{12}\frac{54}{24^b}\Rightarrow b=-5;c=8\Rightarrow P=15$

[Toanhochoctoan]

$\left\{\begin{matrix} \log_712=x & & \\ \log_{12}24=y & & \end{matrix}\right. \Rightarrow xy=\log_712.\log_{12}24=\log_724 \Rightarrow \log_{54}168=\frac{\log_7168}{\log_754}=\frac{\log_724+\log_77}{\log_754}=\frac{xy+1}{\log_754}\Rightarrow a=1$
Suy ra $bxy+cx=\log_754\Leftrightarrow b.\log_754+c.\log_712=\log_754\Leftrightarrow \log_7(24^b.12^c)=\log_754\Leftrightarrow c=\log_{12}\frac{54}{24^b}\Rightarrow b=-5;c=8\Rightarrow P=15$

Bạn ơi. Khúc cuối sao suy ra được $b,c$ vậy.