# Bài 38 trang 93 SGK Đại số và Giải tích 12 Nâng cao

##### Hướng dẫn giải

a) $$\log {1 \over 8} + {1 \over 2}\log 4 + 4\log \sqrt 2 = - \log 8 + \log 2 + \log 4 = - \log 8 + \log 8 = 0$$

b) $$\log {4 \over 9} + {1 \over 2}\log 36 + {3 \over 2}\log {9 \over 2} = \log \left( {{4 \over 9}.6\sqrt {{{\left( {{9 \over 2}} \right)}^3}} } \right) = \log \left( {{4 \over 9}.6.{{{3^3}} \over 2}.\sqrt {{1 \over 2}} } \right)$$

$$= \log \left( {{4 \over 9}{{.3}^4}.{{\sqrt 2 } \over 2}} \right) = \log \left( {18\sqrt 2 } \right)$$

c) $$\log 72 - 2\log {{27} \over {256}} + \log \sqrt {108} = \log \left( {{2^3}{{.3}^2}} \right) - \log {{{3^6}} \over {{2^{16}}}} + \log \sqrt {{2^2}{{.3}^3}}$$

$$= \log \left( {{2^3}{{.3}^2}:{{{3^6}} \over {{2^{16}}}}{{.2.3}^{{3 \over 2}}}} \right) = \log \left( {{2^{20}}{{.3}^{ - {5 \over 2}}}} \right) = 20\log 2 - {5 \over 2}\log 3$$.

d) $$\log {1 \over 8} - \log 0,375 + 2\log \sqrt {0,5625} = \log {2^{ - 3}} - \log \left( {0,{5^3}.3} \right) + \log \left( {0,{5^4}{{.3}^2}} \right)$$

$$= \log {2^{ - 3}} - \log {2^{ - 3}} - \log 3 + 2\log {2^{ - 2}} + 2\log 3 = \log {2^{ - 4}} + \log 3 = \log {3 \over {16}}$$.