log(9)x+log(x^2)3=1的解为?原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/19 01:33:22
log(9)x+log(x^2)3=1的解为?原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,

log(9)x+log(x^2)3=1的解为?原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,
log(9)x+log(x^2)3=1的解为?
原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0
请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,感谢ing……

log(9)x+log(x^2)3=1的解为?原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,
因为log(x)3与log(3)X是互为倒数关系,
所以,在方程1/2*log(3)X+1/2*log(x)3=1两边同乘1/2*log(x)3,
即可得到:[log(3)x]²+1=2log(3)x ,
移项,得:[log(3)x]²-2log(3)x +1=0,
这里:[log(3)x]²-2log(x)3+1=0,——你打错了.

也可以这样:log(3)x+log(x)3=2
log(3)x+log(x)3=log(3)3+log(x)x
log(3)x-log(3)3=log(x)x-log(x)3
log(3)(x-3)=log(x)(x-3)
x=3你好,谢谢你的答案,但我想问下。如果log(3)x-log(3)3=log(x)x-log(x)3有 log(3)(x-3)=log(x)...

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也可以这样:log(3)x+log(x)3=2
log(3)x+log(x)3=log(3)3+log(x)x
log(3)x-log(3)3=log(x)x-log(x)3
log(3)(x-3)=log(x)(x-3)
x=3

收起

此题用均值不等式来解最好:log(9)x+log(x^2)3=[log(3)x+log(x)3]/2=1,log(3)x+log(x)3=2,因为log(3)x与log(x)3互为倒数,所以log(3)x与log(x)3同号,显然依题意有log(3)x与log(x)3同为正数,log(3)x+log(x)3>=2当log(3)x=log(x)3,即log(3)x=log(x)3=1,此为x=3

333333

不对啊