Tính các biểu thức sau: 

1)  $\log _a\sqrt[3]{\sqrt{a} } $

2)  $\frac{1}{2}\log _736-\log _714-3\log _7\sqrt[3]{21} $

3)  $\log _5\frac{1}{25}.\log_{27}9$

4)  $2\log _{27}.9\log 1000$
$\log_5 \dfrac{1}{25} = \log_5 5^{-2} = - 2 \ ,\log_{27} 9 = \log_{3^3} 3^2 = \dfrac{2}{3}\log_3 3 = \dfrac{2}{3}$ 

Kq $ = -2 + \dfrac{2}{3} = -\dfrac{4}{3}$
$\dfrac{1}{2}\log_7 36 = \dfrac{1}{2}\log_7 6^2 = \log_7 6$

$\log_7 14 = \log_7 2 + \log_7 7 = \log_7 2 + 1$

$3\log_7 \sqrt[3]{21} = 3\log_7 21^{\frac{1}{3}} = \log_7 21 = \log_7 3 + \log_7 7 = \log_7 3 + 1$

Kq $= \log_7 6 - \log_7 2 - \log_7 3 - 2 = \log_7 6 - (\log_7 2 + \log_7 3) - 2 = \log_7 6 - \log_7 6 - 2 = 2$
uhm - 2 :D –  Dép Lê Con Nhà Quê 27-07-13 11:57 AM
= - 2 chứ bạn nhỉ –  mackhue59 27-07-13 11:11 AM
1) Điều kiện $0 < a \ne 1$

$\log_a \sqrt[3]{\sqrt a} = \log_a \sqrt[6]{a} = \log_a a^{\frac{1}{6}} = \dfrac{1}{6}$
3. $\log_5\dfrac{1}{25}.\log_{27}9$
$=-2.\dfrac{2}{3}=\dfrac{-4}{3}$
$2\log_{27} 9 = \dfrac{4}{3}$ tính giống câu trên

$log 1000 = \log 10^3 = 3\log 10 = 3$

KQ $= 3 .9.\dfrac{4}{3} = 36$
thi nhân thêm 9 vào, chài ah thắc mắc chi ^^ –  Dép Lê Con Nhà Quê 27-07-13 11:53 AM
bạn ơi câu này phải là 2 log 27 nhân với 9 log 1000 cơ –  mackhue59 27-07-13 11:14 AM

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