Chủ đề này có 2 trả lời

[ntnt]

Hãy tính:

  a) $\log_616,$ biết: $\log_{12}27=a$

  b) $\log_635,$ biết:  $\left\{ \begin{array}{l}\log_{27}5=a\\\log_87=b\\\log_23=c\end{array} \right.$

  c) $\log_a\dfrac{a^2\sqrt[4]{b}\times c^2}{\sqrt[3]{a}\times b^4\sqrt[]{c}},$ biết: $\left\{ \begin{array}{l}\log_ab=3\\ \log_ac=-2 \end{array} \right.$

 

 

[chanhquocnghiem]

a

 

 

Hãy tính:

  a) $\log_616,$ biết: $\log_{12}27=a$

  b) $\log_635,$ biết:  $\left\{ \begin{array}{l}\log_{27}5=a\\\log_87=b\\\log_23=c\end{array} \right.$

  c) $\log_a\dfrac{a^2\sqrt[4]{b}\times c^2}{\sqrt[3]{a}\times b^4\sqrt[]{c}},$ biết: $\left\{ \begin{array}{l}\log_ab=3\\ \log_ac=-2 \end{array} \right.$

 

a) $log_{12}27=a$ ---> $log_{12}3=\frac{a}{3}$ ---> $log_{12}4=log_{12}(\frac{12}{3})=1-\frac{a}{3}$ ---> $log_{12}16=2-\frac{2a}{3}=\frac{6-2a}{3}$

...$log_{12}3=\frac{a}{3}$ ---> $log_{3}12=\frac{3}{a}$ ---> $log_{3}4=\frac{3}{a}-1=\frac{3-a}{a}$ ---> $log_{3}2=\frac{1}{2}log_{3}4=\frac{3-a}{2a}$

...---> $log_{2}3=\frac{2a}{3-a}$ ---> $log_{2}6=log_{2}3+1=\frac{2a}{3-a}+1=\frac{3+a}{3-a}$ ---> $log_{6}2=\frac{3-a}{3+a}$

...---> $log_{6}12=1+log_{6}2=1+\frac{3-a}{3+a}=\frac{6}{3+a}$

...---> $log_{6}16=log_{6}12.log_{12}16=\frac{6}{3+a}.\frac{6-2a}{3}=\frac{12-4a}{3+a}$

 

b) 

$log_{2}3=c$ ---> $log_{2}6=log_{2}2+log_{2}3=1+c$ ---> $log_{6}2=\frac{1}{1+c}$

$log_{8}7=b$ ---> ---> $log_{2}7=3b$

$log_{27}5=a$ ---> $log_{3}5=3a$ ---> $log_{2}5=log_{2}3.log_{3}5=3ac$

---> $log_{6}35=log_{6}2.(log_{2}7+log_{2}5)=\frac{1}{1+c}.(3b+3ac)=\frac{3(b+ac)}{1+c}$

 

c)

$log_{a}\frac{a^2.\sqrt[4]{b}.c^2}{\sqrt[3]{a}.b^4.\sqrt{c}}=log_{a}(a^{\frac{5}{3}}b^{-\frac{15}{4}}c^{\frac{3}{2}})=\frac{5}{3}+(-\frac{15}{4}).3+\frac{3}{2}.(-2)=-\frac{151}{12}$

 

 

[thuy32]

a) ta có : $log_{6}^{16}=\frac{4}{log_{2}^{6}}=\frac{4}{1+log_{2}^{3}}$

* $log_{12}^{27}=a\Leftrightarrow \frac{log_{3}^{27}}{log_{3}^{12}}=a\Leftrightarrow \frac{3}{1+log_{3}^{4}}=a\Leftrightarrow log_{3}^{4}=\frac{3-a}{a}$

$log_{4}^{3}=\frac{a}{3-a}\Leftrightarrow log_{2}^{3}=\frac{2a}{3-a}$

$\Rightarrow log_{6}^{16}=\frac{4(3-a)}{3+a}.$

 

b) Ta có: $log_{6}^{35}=log_{6}^{5}+log_{6}^{7}=\frac{log_{3}^{35}}{log_{3}^{6}}+\frac{log_{2}^{7}}{log_{2}^{6}}=\frac{log_{3}^{35}}{1+log_{3}^{2}}+\frac{log_{2}^{7}}{1+log_{2}^{3}}$

Theo gt:           $\left\{\begin{matrix} log_{27}^{5}=a & \\ log_{8}^{7}=b & \\ log_{2}^{3}=c& \end{matrix}\right.$ $\Leftrightarrow \left\{\begin{matrix} log_{3}^{5}=3a & \\ log_{2}^{7}=3b & \\ log_{3}^{2}=\frac{1}{c}& \end{matrix}\right.$

Vậy $log_{6}^{35}=\frac{3(ac+b)}{1+c}$

c) Ta có: $b=a^{3}; c=a^{-2}$

$\frac{a^{2}\sqrt[4]{b}.c^{2}}{\sqrt[3]{a}.b^{4}.\sqrt[]{c}}$ 

$=\frac{a^{2}.{{(a^{3})}^{\frac{1}{4}}}.({a^{-2}})^{2}}}{{a^\frac{1}{3}.({a^{3}})^{4}.({a^{-2}})^{\frac{1}{2}}}$

$=a^{\frac{-151}{12}}$

Vậy $log_{a}^{\frac{a^{2}\sqrt[4]{b}.c^{2}}{\sqrt[3]{a}.b^{4}.\sqrt[]{c}}}=\frac{-151}{12}$

Bài viết đã được chỉnh sửa nội dung bởi thuy32: 14-10-2013 - 13:40