(log₄3+log₈3)(log₃2+log₉2)=?
解:原式=[(1/2)log₂3+(1/3)log₂3][log₃2+(1/2)log₃2]=[(5/6)log₂3][(3/2)log₃2]=(5/6)(3/2)=5/4
其中(log₂3)(log₃2)=1
你好:
(log4^3+log8^3)(log3^2+log9^2)
=(log2^6+log2^9)(log3^2+log3^4)
=(6log2+9log2)*(2log3+4log3)
=15log2*6log3
=90log2*log3
希望对你有帮助!
(log4^3+log8^3)(log3^2+log9^2)
=(log(2^2)^3+log(2^3)^3)(log3^2+log(3^2)^2)
=(log2^6+log2^9)(log3^2+log3^4)
=(6log2+9log2)(2log3+4log3)
=15log2*6log3
=90log2log3
(log4^3+log8^3)(log3^2+log9^2) 题目lg4^3吧,要不没有底数怎么搞,,
=(log(2^2)^3+log(2^3)^3)(log3^2+log(3^2)^2)
=(log2^6+log2^9)(log3^2+log3^4)
=(6log2+9log2)(2log3+4log3)
=15log2*6log3=90log2log3 应该
(log4^3+log8^3)(log3^2+log9^2)=(3log4+3log4^2)(2log3+2log3^2)=(3log4+6log4)(2log3+4log3)=(9log4)(6log3)=54log4log3=108log2log3