Answer: \(x=\log_3 4\).

Let

\(u=3^x.\)

Then \(3^{2x}=u^2\) and \(3^{x+1}=3u\). The equation becomes

\(u^2-3u-4=0.\)

Factorise:

\((u-4)(u+1)=0.\)

So \(u=4\) or \(u=-1\). Since \(u=3^x\) is positive, reject \(u=-1\).

Therefore

\(3^x=4,\)

and hence

\(\boxed{x=\log_3 4}.\)