log2(根号下7/48)+log2(12)-1/2log2(42)-1=?

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/12 16:39:55

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=? log2 SQR(7/48)+log2 12 -1/2log2 42 log2 SQR(7/48)+log2 12 -1/2log2 48 求化简log2根号下48分之7+log2 12-log2 42 1/2log2(7/48)+log2 12-1/2log2 42 Log2√（7/48） +log2 12 -（1/2）log2 42 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=? 根号下{[log2 （5)]^2-4log2(5)+4}+log2(1/5) 根号下[(log2^3)^2-4log2^3+4]+log2^(1/3) log2[-x+根号下(x^2+1)] f(x)=log2^根号下1+sinx + log2^根号下1-sinx 的定义域 log2（√1/48）+log2 12-1/2log2 （24）-1/3