log2(根号下7/48)+log2(12)-1/2log2(42)-1=?
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log2(根号下7/48)+log2(12)-1/2log2(42)-1=?
log2(根号下7/48)+log2(12)-1/2log2(42)-1=?
log2(根号下7/48)+log2(12)-1/2log2(42)-1=?
log2[根号下(7/48)]+log2(12)-1/2log2(42)-1?
=log2{12*根号下(7/48)/[根号下(42)]}-1
=log2{12*根号下[7/(48*42)]}-1
=log2{12*根号下[7/(6*2*2*2**6*7)]}-1
=log2根号下(1/2)-1
=(1/2)log2(1/2)-1
=(1/2)log2(2^-1)-1
=(-1)(1/2)log2(2)-1
=-1/2-1
=-3/2
log2(根号下7/48)+log2(12)-1/2log2(42)-1=?
求化简log2根号下48分之7+log2 12-log2 42
log2根号7/48+log2^12-1/2log2^42-1=
log2 SQR(7/48)+log2 12 -1/2log2 48
log2 SQR(7/48)+log2 12 -1/2log2 42
根号下[(log2^3)^2-4log2^3+4]+log2^(1/3)
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f(x)=log2^根号下1+sinx + log2^根号下1-sinx 的定义域值域
log2(3)×log2(4)=log2(7)?log2(3)×log2(4)=log2(7)吗?log2(1/49)+log2(1/16)+log2(1/27)等于多少呢?帮个忙!
log2为底数根号7/48为真数+log2为底数12为真数-1/2log2为底数42为真数=?
1/2log2(7/48)+log2 12-1/2log2 42log2 的2 在g的右下脚
化简算式log2^√7/48+log2^12-1/2log2^42+1/2^log2^3
求值:log2√(7/48)+log2(12)-1/2log2(42)-1 这道题我算来算去算不对
计算log2 √(48/7)-log2 12+(1/2)log2 42-1
log2[√(7/48)]+log2(12)-log2(√42)
y=根号下log2(x-1)的定义域