Write 9 = 3² and 27 = 3³, then 1/9 = 3⁻² and 1/27 = 3⁻³.
So, we have
[tex]\left(\dfrac1{27}\right)^{2x+1} = \left(\dfrac19\right)^{x+4}[/tex]
[tex]\left(3^{-3}\right)^{2x+1} = \left(3^{-2}\right)^{x+4}[/tex]
Recall that [tex](a^b)^c = a^{bc}[/tex] for real a, b, and c. Then this equation is the same as
[tex]3^{-3(2x+1)} = 3^{-2(x+4)}[/tex]
[tex]3^{-6x - 3} = 3^{-2x - 8}[/tex]
The bases on either side are the same, so the exponents must be equal:
-6x - 3 = -2x - 8
Solve for x :
-4x = -5
x = 5/4
