已知f(x)=2^x,设f(x)的反函数为f-1(x),若关于x的方程f-1(ax) f-1(ax^2)=f-1(16)
问题描述：
已知f(x)=2^x,设f(x)的反函数为f-1(x),若关于x的方程f-1(ax) f-1(ax^2)=f-1(16)
a=4,球x
x在（0,1）球a
答
x=log2(x)
f-1(ax) f-1(ax^2)=f-1(16)
中间是+还是-f逆ax 乘 f逆ax^2=f逆16log2(ax)*log2(ax^2)=log2(16)=4(1)a=4log2(4x)*log2(4x^2)=4[log2(4)+log2(x)]*[log2(4)+log2(x^2)]=4[2+log2(x)][2+2log2(x)]=4设log2(x)=t(t+1)(t+2)=2t^2-3t=0t=0或t=3log2(x)=0x=1log2(x)=3x=2^3=8(2)稍等不对啊是（t+2）（t+2）20 0无解(t+1)(t+2)=2t^2+3t+2=2t^2+3t=0t=-3或t=0log2(x)=0x=1log2(x)=-3x=2^(-3)=1/8(2)[log2(a)+log2(x)]*[log2(a)+log2(x^2)]=4[log2(a)+log2(x)]*[log2(a)+2log2(x)]=4设log2(a)=mlog2(x)=n∵0