3^(1+log3 6) -2^(4+log2 3)+10^3lg3+(9/1)^log3^4
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3^(1+log3 6) -2^(4+log2 3)+10^3lg3+(9/1)^log3^4
题有个错误,+(9/1)^log3^4改为+(9/1)^log3^4-1
题有个错误,+(9/1)^log3^4改为+(9/1)^log3^4-1
原式=3*3^log3(6)-(2^4)*2^log2(3)+(10^lg3)^3+(3^-2)^log3(4)
=3*6-16*3+3^3+[3^log3(4)]^(-2)
=18-48+27+4^(-2)
=-5+1/16
=-79/16
再问: 题有个错误,+(9/1)^log3^4改为+(9/1)^log3^4-1
再答: -1是在指数上的吧? 原式=3*3^log3(6)-(2^4)*2^log2(3)+(10^lg3)^3+[(3^-2)^log3(4)]/(1/9) =3*6-16*3+3^3+9[3^log3(4)]^(-2) =18-48+27+9*4^(-2) =-5+9/16 =-71/16
再问: 3*6-16*3+3^3+9[3^log3(4)]^(-2) 这一步里面的3*6,16*3,3^3,9[3^log3(4)]^(-2)是怎么得到的,还是有点不懂。。。
再答: a^loga(n)=n 所以:3^log3(6)=6,同理:2^log2(3)=3,(10^lg3)=3
=3*6-16*3+3^3+[3^log3(4)]^(-2)
=18-48+27+4^(-2)
=-5+1/16
=-79/16
再问: 题有个错误,+(9/1)^log3^4改为+(9/1)^log3^4-1
再答: -1是在指数上的吧? 原式=3*3^log3(6)-(2^4)*2^log2(3)+(10^lg3)^3+[(3^-2)^log3(4)]/(1/9) =3*6-16*3+3^3+9[3^log3(4)]^(-2) =18-48+27+9*4^(-2) =-5+9/16 =-71/16
再问: 3*6-16*3+3^3+9[3^log3(4)]^(-2) 这一步里面的3*6,16*3,3^3,9[3^log3(4)]^(-2)是怎么得到的,还是有点不懂。。。
再答: a^loga(n)=n 所以:3^log3(6)=6,同理:2^log2(3)=3,(10^lg3)=3
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