log(x+2)^(4x+5)-log(4x+5)^(x^2+4x+4)=1

如题解答，请高手写出步骤

在此谢谢

用换底公式x+2＞0，4x+5＞0得x＞-5/4

原式=lg(4x+5)/lg(x+2)-lg(x+2)^2/lg(4x+5)=1

两边同乘以lg(x+2)lg(4x+5)得

[lg(4x+5)]^2-2[lg(x+2)]^2-lg(4x+5)lg(x+2)=0

得[lg(4x+5)-2lg(x+2)]×[lg(4x+5)+lg(x+2)]=0

得4x+5=(x+2)^2或4x+5=-(x+2)

得x=±1或x=-7/5＜-5/4，舍去

所以x=±1

可化为log(4x+5)/log(x+2)-log(x+2)^2/log(4x+5)=1 log^2(4x+5)-2log^2(x+2)=log(x+2)log(4x+5) [log(4x+5)-2log(x+2)][log(4x+5)+log(x+2)]=0 log(4x+5)-2log(x+2)=0 4x+5=(x+2)平方 x平方=1 x=±1 又x+2和4x+5≠1 所以x=1