解方程:(1)log2^x log2^(x-1)=1

log2^(4^x+4)=x+log2^[2^(x+1)-3]log2^(4^x+4)-log2^[2^(x+1)-3]=xlog2[(4^x+4)/(2^(x+1)-3)]=x(4^x+4)/(2^

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

则有log2(2^x+1)=正负根号下2所以当log2(2^x+1)=根号下2时.2^x+1=2^根号下2时则x=log2[（2^根号下2时）-1]同理为log2(2^x+1)负根号下2时x=log2

先换成同底的log4(2-x)=log4(x-1）^2-log44化简一下log4(2-x)=log4[(x-1）^2/4]然后2-x=（x-1）^2/4结果是x=2或-42舍掉（因为真数是0）-4也

log根号2(x+1)-log2(x+5/2)=1log2(x+1)²-log2(x+5/2)=1log2(x+1)²/(x+5/2)=1所以(x+1)²/(x+5/2)

log2（9x-1-5）=log2（3x-1-2）-2．∴log2（9x-1-5）=log2（3x-1-2）-log24．∴log2（9x-1-5）=log24（3x-1-2）．∴9x-1-5=4（3

log2(2-x)=log2(x-1)+log(2)log2(2-x)=log2[2(x-1)]2-x=2x-2x=4/3带入检验,成立所以x=4/3

具体过程嘛.我写关键步成吗.log2(2((2^x)+1)))=2/log2((2^x)+1)log2(2)+log2((2^x)+1)=2/log2((2^x)+1)1+log2((2^x)+1)=

log2(x+14)-log2(x+6)=3-log2(x+2)∴log2[(x+14)/(x+6)]=log2[8/(x+2)]∴(x+14)/(x+6)=8/(x+2)且(x+14)/(x+6)>

由题意定义域2x+1>0x^2-2>0则x>√2而x^2-2=2x+1则x=3或x=-1则x=3

求定义域,x

由x+4>0,x-1>0,x+8>0得x>1∵log2(x+4)+log2(x-1)=log2(2)+log2(x+8)∴log2(x+4)(x-1)=log2(x+8)*2,即(x+4)(x-1)=

log2(4^x+4)=log2(2^x)+log2[2^(x+1)-3]令2^x=a4^x=a²2^(x+1)=2alog2(a²+4)=log2(a)+log2(2a-3)lo

log2(3-x)-log4(x+5)=1log2(3-x)=1+log4(x+5)log4(3-x)²=log44(x+5)(3-x)²=4(x+5)x²-6x+9=4

log2[9^(x-2)+7]-log2[3^(x-1)+1]=2即log2[（9x-11）/（3x-2）]=2（9x-11）/（3x-2）=4所以x=-1但又因9^(x-2)+7大于0所以x大于11

由题意可知：令t＝log(2x+1)2，则t（t+1）=2，所以t=1或-2．由log2（2x+1）=1，可知x=0；由log2（2x+1）=-2，可知无解；所以方程的解为0．故答案为：0．

log2(4^x+4)=log2(2^x)+log2[2^(x+1)-3]令2^x=a4^x=a²2^(x+1)=2alog2(a²+4)=log2(a)+log2(2a-3)lo

log2（9^（x-1）-5）-log2（3^x+2）-2=0log2（9^（x-1）-5）=log2（3^x+2）+log2(4)log2（9^（x-1）-5）=log2（3^x*4+8）9^（x-

题目是这样的吗?后面是不是lg?如果是,先换底.然后……

∵log2（x-1）=2-log2（x+1）∴log2（x-1）=log24x+1即x-1=4x+1解得x=±5（负值舍去）故答案为：5