(1/27)x+2+1=(1/3)x-3+1
\(\color{blue}{\begin{array}{rll}(\dfrac{1}{27})^{x+2}+1&=&(\dfrac{1}{3})^{x-3}+1\\ (3^{-3})^{x+2}&=&(3^{-1})^{x-3}\\ 3^{-3x-6}&=&3^{3-x}\\ -3x - 6&=& 3-x\\ -2x&=&9\\ x&=&-\dfrac{9}{2}\end{array}}\)
#BlackLaTeXIsBoring.
(1/27) x+2 +1=(1/3) x-3 +1
(Headline of the task)
\(\frac{1}{27}x+2+1=\frac{1}{3}x-3+1\)
\((\frac{1}{3}-\frac{1}{27})x=2+1+3-1\)
\(\frac{8}{27}x=5\)
\(x=\frac{5\times 27}{8}\)
\(x=16\frac{7}{8}\) !
!
