解方程log2(4^x+1)=x+log2(2^(x+3)-6)

显示全部楼层 · 2011-3-19 18:18:39

你好log2(4^x+1)=x+log2[2^(x+3)-6]移项log2(4^x+1)-log2[2^(x+3)-6]=xlog2[(4^x+1)/(2^x×8-6)]=x即2^x=(4^x+1)/(2^x×8-6)去分母2^x(2^x×8-6）=4^x+18×4^x-6×2^x=4^x+17×4^x-6×2^x-1=0令2^x=t,则t＞07t2-6t-1=0解得t=1或t=-1/7(舍去）所以2^x=1x=0所以原方程的解为x=0

千问 · 2011-3-19 18:18:39

log2(4^x+1)=log2(2^x)+log2(2^(x+3)-6)log2(4^x+1)=log2((2^x)(2^(x+3)-6))4^x+1=(2^x)(2^(x+3)-6)2^(2x)+1=(2^x)(8(2^x)-6)射2^x=yy^2+1=y(8y-6)y^2+1=8y^2-6y7y^2-6y-1=0y=1