7lg20•(
12
)lg0.7
=
14
14
分析:根據(jù)對數(shù)的運算性質(zhì),我們可以得到algb=blga,進(jìn)而可將原式化為20lg7(
1
2
)
lg0.7
,再由指數(shù)的運算性質(zhì)和對數(shù)的運算性質(zhì)化簡式子,可得答案.
解答:解:∵lga•lgb=lgb•lga
∴l(xiāng)g(algb)=lg(blga
即algb=blga
7lg20(
1
2
)
lg0.7

=20lg7(
1
2
)
lg0.7

=2lg7•10lg7(
1
2
)
lg0.7

=7•2lg7•2-lg0.7
=7•2lg7-lg0.7
=7•2lg10
=7•2
=14
故答案為14
點評:本題考查的知識點是對數(shù)的運算性質(zhì),指數(shù)的運算性質(zhì),其中由對數(shù)的運算性質(zhì)推出algb=blga,是解答本題的關(guān)鍵.
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