求值:log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=log2中2为底数=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的?
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![求值:log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=log2中2为底数=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的?](/uploads/image/z/3882293-53-3.jpg?t=%E6%B1%82%E5%80%BC%EF%BC%9Alog2+%285%2F56%29%5E%281%2F2%29%2Blog2+%2814%29-%281%2F2%29log2+%2835%29%3Dlog2%E4%B8%AD2%E4%B8%BA%E5%BA%95%E6%95%B0%3D1%2F2%5B-2log2+%287%29-3%5D%2Blog2+%282%29%2Blog2+%287%29+%E4%B8%AD%E7%9A%841%2F2%5B-2log2+%287%29-3%5D%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0%E7%9A%84%EF%BC%9F)
求值:log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=log2中2为底数=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的?
求值:log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=
log2中2为底数
=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的?
求值:log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)=log2中2为底数=1/2[-2log2 (7)-3]+log2 (2)+log2 (7) 中的1/2[-2log2 (7)-3]是怎么得到的?
既然问道我了,就再写详细些:
=1/2(log2 [7^(-2)*2^(-3)]+log2(2*7)
=1/2[(log2 7^(-2)+log2 2^(-3)]+log2 (2*7)
=1/2log2 7^(-2)+1/2log2 2^(-3)]+log2 (2*7)
=1/2*2*log2 7+1/2*(-3)log2 2+log2 (2*7)
=1/2[-2log2 (7)-3]+log2 (2)+log2 (7)
log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)
=1/2log2 (5/56)-1/2log2 (35)+log2 (14)
=1/2[log2 (5/56/35)]+log2 (14)
=1/2log2 [1/(7^2*8)]+log2 (14)
=1/2(log2 [7^(-2)*2^(-3)]+log2(2*7)
=1/2[-2log2 (7)-3]+log2 (2)+log2 (7)
=-log2 (7)-3/2+1+log2 (7)
=-1/2
log2 (5/56)^(1/2)+log2 (14)-(1/2)log2 (35)
=-1/2
mi