Answer:
[tex]a=-\frac{11}{3}[/tex]
Step-by-step explanation:
We want to find the value of [tex]a[/tex] for which [tex]9=(\frac{1}{27})^{a+3}[/tex].
We rewrite with a base of 3 to get:
[tex]3^2=(\frac{1}{3^3})^{a+3}[/tex].
Recall that: [tex]\frac{1}{a^n}=a^{-n}[/tex]
[tex]\implies 3^2=3^{-3(a+3)}[/tex].
We now equate the exponents to get:
[tex]2=-3(a+3)[/tex]
[tex]2=-3a-9[/tex]
[tex]2+9=-3a[/tex]
[tex]11=-3a[/tex]
[tex]a=-\frac{11}{3}[/tex]
The first choice is correct
