solving for x

3^(2(x+1)-8(3^(x+1))=9

\(3^{[2(x+1)-8*3^{(x+1)}]}=9\\ 3^{[2(x+1)-8*3^{(x+1)}]}=3^2\\ 2(x+1)-8*3^{(x+1)}=2\\ 2[(x+1)-4*3^{(x+1)}]=2\\ (x+1)-4*3^{(x+1)}=1\\ x+1-4*3^{(x+1)}=1\\ x-4*3^{(x+1)}=0\\ x=4*3^{(x+1)}\\ \frac{x}{4}=3^{(x+1)}\\ log(\frac{x}{4})=log(3^{(x+1)})\\ log(x)-log(4)=(x+1)log(3)\\ log(x)-log(4)=xlog(3)+log(3)\\ log(x)-log(4)=log(3^x)+log(3)\\ log(x)-log(3^x)=log(3)+log(4)\\ log(\frac{x}{3^x})=log(12)\\ \frac{x}{3^x}=12\\\)

And that is about as far as I can go  :/

I just realised that your brackets are not paired so you probably meant a different problem   

If this is meant to be 3^(2[x+1]) - 8*3^(x+1) = 9 then:

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