Use the binomial theorem:
[tex](a+b)^n=\displaystyle\sum_{k=0}^n\binom nka^{n-k}b^k[/tex]
In this case,
[tex](3x+4)^{11}=\displaystyle\sum_{k=0}^{11}\binom{11}k(3x)^{11-k}4^k[/tex]
The [tex]x^7[/tex] term occurs for [tex]k=4[/tex]:
[tex]\dbinom{11}4(3x)^{11-4}4^4=\dfrac{11!}{4!(11-4)!}3^74^4x^7=184,757,760x^7[/tex]
