Solve
6^x+1 = 4^2x + 1
start by subtracting 1 from both sides and taking the log
$\displaystyle 6^x=4^{2x} \implies \ln(6^x)=\ln(4^{2x}) \implies x\ln(6)=2x\ln(4)$
Collect your x's and factor
$\displaystyle x\ln(6)-2x\ln(4)=0 \implies x(ln(6)-2ln(4))=0$
Now by the zero factor theorem we have x=0.
Good luck