This series is geometric; substitute [tex]y=4^x-7[/tex] to see how.
Then the series converges if
[tex]|4^x-7|<1\implies-1<4^x-7<1\implies6<4^x<8\implies\log_46<x<\log_48[/tex]
Under these conditions, we have
[tex]\displaystyle\sum_{n=0}^\infty(4^x-7)^n=\frac1{1-(4^x-7)}=\frac1{8-4^x}[/tex]
