2^log4(2-根号3)^2+3^log(2+根号3)^2[(1-log6(3)^2+log6(2)*log6(18)]/log6(4)[(1-log6(3))^2+log6(2)*log6(18)]/log6(4)

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2^log4(2-根号3)^2+3^log(2+根号3)^2[(1-log6(3)^2+log6(2)*log6(18)]/log6(4)[(1-log6(3))^2+log6(2)*log6(18)]/log6(4)
2^log4(2-根号3)^2+3^log(2+根号3)^2
[(1-log6(3)^2+log6(2)*log6(18)]/log6(4)
[(1-log6(3))^2+log6(2)*log6(18)]/log6(4)

2^log4(2-根号3)^2+3^log(2+根号3)^2[(1-log6(3)^2+log6(2)*log6(18)]/log6(4)[(1-log6(3))^2+log6(2)*log6(18)]/log6(4)
2^log4(2-√3)^2+3^log9(2+√3)^2
=[4^(1/2)]^log4(2-√3)^2+[9^(1/2)]^log9(2+√3)^2
=4^[(1/2)log4(2-√3)^2]+9^[(1/2)*log9(2+√3)^2]
=4^log4(2-√3)^[2*(1/2)]+9^log9(2+√3)^[2*(1/2)]
=4^log4(2-√3)+9^log9(2+√3)
=(2-√3)+(2+√3)
=4
{[1-log6(3)]^2+log6(2)*log6(18)}/log6(4)
={[log6(6)-log6(3)]^2+log6(2)*log6(18)}/log6(4)
={[log6(6/3)]^2+log6(2)*log6(18)}/log6(4)
={[log6(2)]^2+log6(2)*log6(18)}/log6(4)
={log6(2)*[log6(2)+log6(18)]}/log6(4)
=[log6(2)*log6(2*18)]/log6(4)
=[log6(2)*log6(36)]/log6(4)
=[log6(2)*log6(6^2)]/log6(4)
=[log6(2)*2*log6(6)]/log6(4)
=2log6(2)/log6(4)
=log6(2^2)/log6(4)
=log6(4)/log6(4)
=1

log4^(1+根号2+根号3）+log4^（1+根号2-根号3）呵呵,摆脱了,呵呵, (log4(3)+log8(3))(log3(2)+log9(2))-log4(3)×log1/3(四次根号32) 2^log4(3) log4(3x-2) 2^log4^3=? 计算log9(根号2)*[log4(3)+log8(3)] (log2 5+log4 125)log3 2/ log根号3 5 2log4 (2-根号3)+3log9(2+根号3)错了 是2^(log4 (2-根号3))+3^(log9(2+根号3)) 计算：2^[log4(3) -2] 计算：2^log4(2--根号3）^2+2^log(2+根号3）^2 2^(log4 (2-根号3))+3^(log9(2+根号3))= 2log4 (2-根号3)²+3log9(2+根号3)² log4的根号2次幂是多少? log4(3)*log9(2)+log2(4倍根号下64) log9(32)*log64(2log4(3)*log9(2)+log2(4倍根号下64) log9(32)*log64(27)+log9(2)*log4(根27) (log3,4+log9,4)log4,3—log2,2根号2 这怎么解答了 对数函数计算 log4 3·log9 2-log1/2 4次根号32 (log2(5）+log4（125））×log3（2）/log根号3（5） (log2^5+log4^125)(log3^2)/(log(根号3)^5 要具体的过程,