Помогите решить неравенство 6.50 (а), пожалуйста

Решите задачу:

$3^{x+4}+3^{-x-3}\ \textgreater \ 4\\ 3^x=m, \ m\ \textgreater \ 0\\ 3^4*m +\frac{1}{3^3*m} \ \textgreater \ 4\\ 81m+ \frac{1}{27m} \ \textgreater \ 4/*27m\\ 2187m^2+1-108m\ \textgreater \ 0\\ 2187m^2-108m+1\ \textgreater \ 0\\ D=(-108)^2-4*2187=54^2\\ m_1= \frac{108-54}{2*2187} = \frac{1}{81} \\ m_2= \frac{108+54}{2*2187} = \frac{1}{27} \\ (m- \frac{1}{81} )(m- \frac{1}{27} )\ \textgreater \ 0\\ -\infty\ \textless \ m\ \textless \ \frac{1}{27} , \ \frac{1}{81} \ \textless \ m\ \textless \ +\infty\\ 3^x\ \textless \ 3^{-3}, \ 3^x\ \textgreater \ 3^{-4}\\ x\ \textless \ -3, \ x\ \textgreater \ -4\\ x\in(-\infty;-4)U(-3;+\infty)$