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Nimfa-mama [501]
3 years ago
6

For what value of a does 9 = (StartFraction 1 Over 27 EndFraction) Superscript a + 3?

Mathematics
2 answers:
umka2103 [35]3 years ago
7 0

Answer:

a=-\frac{11}{3}

Step-by-step explanation:

We want to find the value of a for which 9=(\frac{1}{27})^{a+3}.

We rewrite with a base of 3 to get:

3^2=(\frac{1}{3^3})^{a+3}.

Recall that: \frac{1}{a^n}=a^{-n}

\implies 3^2=3^{-3(a+3)}.

We now equate the exponents to get:

2=-3(a+3)

2=-3a-9

2+9=-3a

11=-3a

a=-\frac{11}{3}

The first choice is correct

Rus_ich [418]3 years ago
7 0

Answer:

A.-11/3

Step-by-step explanation:

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