(1)f(x)=ln(4^x+1)/ln2-x,
f'(x)=1/[(4^x+1)ln2]*(4^x*ln4)-1=2*4^x/(4^x+1)-1=(4^x-1)/(4^x+1),
x>0时4^x>1,f'(x)>0,
∴f(x)在[0,+∞）上是增函数.
(2)u=a*2^x-4a/3(a>0)是增函数,
由u>0得2^x>4/3,x>log(4/3).
由f(x)=g(x)得logu-log(4^x+1)=x,
log[u/(4^x+1)]=x,设t=2^x,则u/(4^x+1)=a(t-4/3)/(t^2+1)=t>4/3,
∴a=(t^3+t)/(t-4/3)(t>4/3),
∴a'=[(3t^2+1)(t-4/3)-(t^3+t)]/(t-4/3)^2
=(2t^3-4t^2-4/3)/(t-4/3)^2,繁!
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