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,
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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