Answer: \(x=\log_7 11=\frac{\ln11}{\ln7}\).

Set \(y=30\):

\(2(7^{2x})-3(7^{x+1})+19=30.\)

Let \(u=7^x\). Then \(7^{2x}=u^2\) and \(7^{x+1}=7u\).

The equation becomes

\(2u^2-21u+19=30.\)

So

\(2u^2-21u-11=0.\)

Factorise:

\((2u+1)(u-11)=0.\)

Since \(u=7^x\gt 0\), reject \(u=-\frac12\). Hence

\(7^x=11.\)

Therefore

\(x=\log_7 11=\frac{\ln11}{\ln7}.\)