We have to solve for x:
[tex]5(3^{2x+1})-19=6[/tex]
We first isolate the term 3^(2x+1) as:
[tex]\begin{gathered} 5(3^{2x+1})-19=6 \\ 5(3^{2x+1})=6+19 \\ 5(3^{2x+1})=25 \\ 3^{2x+1}=\frac{25}{5} \\ 3^{2x+1}=5 \end{gathered}[/tex]
We can now apply logarithm with base 3 to both sides of the equation and get:
[tex]\begin{gathered} \log _3(3^{2x+1})=\log _35 \\ 2x+1=\log _35 \\ 2x=\log _35-1 \\ x=\frac{\log _35-1}{2} \\ x\approx\frac{1.46497-1}{2} \\ x\approx\frac{0.46497}{2} \\ x\approx0.232 \end{gathered}[/tex]
Answer: x = 0.232
