Question

log2为底(x+1)<=log4为底(3x+1) 怎么解。。急~

Answer 1

log4为底(3x+1) =1/2 log2为底(3x+1) =log2为底(3x+1)的1/2次方，则原不等式可化为
x+1<=(3x+1)^1/2 ,两面同时平方并化简可得x^2-x<=0，即0<=x<=1,同时综合x+1>0,3x+1>0
可得最后结果为0<=x<=1

Answer 2

log2[x+1]≤log4[3x+1]
定义： x+1>0 x>-1
3x+1>0 x>-1/3
log2[x+1]≤(1/2)log2[3x+1]
2log2[x+1]≤log2[3x+1]
log2[x+1]²≤log2[3x+1]
(x+1)²≤3x+1
x²+2x+1≤3x+1
x²-x≤0
x(x-1)≤0
得
0≤x≤1