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Given: 9=(1/27)^(a+3)

[Note:  Please review rules of PEMDAS.  
The question posted as is means
9=[(1/27)^a]+3 which is probably not what you mean.  It has been interpreted with (a+3) as the exponent, please check.]

Need to find value of a that satisfies above.

We will first isolate a in order to solve for its value.
9=(1/27)^(a+3)
Switch unknown to the left
(1/27)^(a+3) = 9
Since both 27 and 9 are powers of 3, we can take log to base 3.
take log (base 3) on both sides, applying rule of logarithm of exponents.
(a+3)log_3(1/27)=log_3(9)
(a+3)log_3(3^(-3))=log_3(3^2)
Apply definition of logarithm [log_a(a^k)=k]
(a+3)(-3)=2
(a+3)=-2/3
a=-3 2/3=-11/3

Check: (1/27)^(-11/3+3)=(1/27)^(-2/3)=(3^(-3))^(-2/3)=3^(-3(-2/3))=3^2=9 ok

Answer: a=-11/3 or a=-3.667 (approx.)

The value of a in the equation [tex]9=(\frac{1}{27})^{a+3}[/tex] is a = -11/3

Solving indical equations

The given equation is:

[tex]9=(\frac{1}{27})^{a+3}[/tex]

This can be rewritten as:

[tex]9=27^{-1(a+3)}[/tex]

Which can be further simplified as:

[tex]3^2=3^{-3(a+3)}[/tex]

Since the bases are equal, they cancel out

The equation becomes:

2  =  -3(a  +  3)

2  =  -3a  -  9

2+9  =  -3a

-3a  =  11

a  =  -11/3

Therefore, the value of a in the equation [tex]9=(\frac{1}{27})^{a+3}[/tex] is a = -11/3

Learn more on indices here: https://brainly.com/question/8952483

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