【求助】高一 一小题对数方程！！！

0.5*x=x的1/2次幂
求解得：x=4或者0，又因为 x>0
所以
x=4

log2[log3(log9 x)] = 2log4[log9(log3 x)] = (2/2)·log2[log9(log3 x)]
∴[log3(log9 x)] = [log9(log3 x)] = (1/2)·[log3(log3 x)]
= log3[(log3 x)^(1/2)]
∴log9 x = (log3 x)^(1/2) = (1/2)·log3 x ，令t = log3 x ，
则：t = t^2/4 ，解得t = 0 或 4 ，解得x = 1 或 81

运用公式log《a^n》(b^m)=(m/n)log《a》(b) 书名号内是底数，小括号内是真数。
上面的公式很容易证明，以a为底，换底就可以到证，证明过程如下：
log《a^n》(b^m)
=[log《a》(b^m)]/[log《a》(a^n)]
=[m*log《a》(b)]/n
=(m/n)*log《a》(b)

上面的公式可以引申出如下变形：
当m=1时，log《a^n》(b)=(1/n)*log《a》(b)
当n=1时，log《a》(b^m)=m*log《a》(b)
m*log《a》(b)=log《a》(b^m)

回到题中来，底数4和9可分别写成2²和3²
log2[log3(log9 x)]=2log4[log9(log3 x)]，化简：
log2[log3(log9 x)]=2log《2²》[log9(log3 x)]
log2[log3(log9 x)]=(2/2)*log《2》[log9(log3 x)]
log2[log3(log9 x)]=log《2》[log9(log3 x)]，去掉外层的log得
log3(log9 x)=log9(log3 x)，再化简
log3(log9 x)=log《3²》(log3 x)
log3(log9 x)=(1/2)