作业帮求值.(log3^2+log9^3)*(log4^3+log8^3)+(log3^3^1/2)^2+ln根号下e-lg1
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求值.(log3^2+log9^3)*(log4^3+log8^3)+(log3^3^1/2)^2+ln根号下e-lg1
我实在是算不好了,要命啊急滴,
我实在是算不好了,要命啊急滴,
题目请描述规范、准确.如,log3^2应该是log9,而不是（log3）^2吧?那么log3^3^1/2是不是log[27^(1/2)]?还有ln根号下e-lg1,指的是ln[(e-lg1)^(1/2)]呢,还是ln[e^(1/2)]-lg1?不说清楚结果是不一样的.
再问： log以3为底2为真数 ln根号下e再减去lg1 （ loglog3^2+log9^3)*(log4^3+log8^3)+1/4+ln根号下e再减去lg1 快！帮帮忙!
再答： 此题需用换底公式。 [log(3)(2)+log(9)(3)]*[log(4)(3)+log(8)(3)] =[1/log(2)(3)+1/2]*[1/2*log(2)(3)+1/3*log(2)(3)] =[1/log(2)(3)+1/2]*5/6*log(2)(3) =5/6+5/12*log(2)(3) 另ln√e=1/2，lg1=0
再问： 对不起啊，题目打错了。（log3^2+log9^2)*(log4^3+log8^3)+1/4+ln根号下e再减去lg1.对不起哈！
再答： （log3^2+log9^2)*(log4^3+log8^3)+1/4+ln√e-lg1 =(log3^2+1/2*log3^2)*[1/2*log2^3+1/3*log2^3]+1/4+1/2-0 =3/2*log3^2*5/6*log2^3+3/4 =5/4+3/4 =2
再问： log以3为底2为真数 ln根号下e再减去lg1 （ loglog3^2+log9^3)*(log4^3+log8^3)+1/4+ln根号下e再减去lg1 快！帮帮忙!
再答： 此题需用换底公式。 [log(3)(2)+log(9)(3)]*[log(4)(3)+log(8)(3)] =[1/log(2)(3)+1/2]*[1/2*log(2)(3)+1/3*log(2)(3)] =[1/log(2)(3)+1/2]*5/6*log(2)(3) =5/6+5/12*log(2)(3) 另ln√e=1/2，lg1=0
再问： 对不起啊，题目打错了。（log3^2+log9^2)*(log4^3+log8^3)+1/4+ln根号下e再减去lg1.对不起哈！
再答： （log3^2+log9^2)*(log4^3+log8^3)+1/4+ln√e-lg1 =(log3^2+1/2*log3^2)*[1/2*log2^3+1/3*log2^3]+1/4+1/2-0 =3/2*log3^2*5/6*log2^3+3/4 =5/4+3/4 =2
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