(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)

$\frac{1}{27}x+2+1=\frac{1}{3}x-3+1$

$(\frac{1}{3}-\frac{1}{27})x=2+1+3-1$

$\frac{8}{27}x=5$

$x=\frac{5\times 27}{8}$

$x=16\frac{7}{8}$      !