Question

algebra: concepts and connections - plc solving exponential equations by rewriting the base for what value of a does 9 = \left(\frac{1}{27}\
ight)^{a + 3}? options: \frac{7}{3}, -\frac{7}{3}, \frac{11}{3}, -\frac{11}{3}

Explanation:

Step1: Rewrite bases as powers of 3

$9 = 3^2$, $\frac{1}{27}=3^{-3}$
So the equation becomes: $3^2 = (3^{-3})^{a+3}$

Step2: Simplify the right-hand side

Use exponent rule $(x^m)^n=x^{m \cdot n}$:
$3^2 = 3^{-3(a+3)}$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$2 = -3(a+3)$

Step4: Solve for $a$

First expand the right side:
$2 = -3a -9$
Add 9 to both sides:
$2+9 = -3a$
$11 = -3a$
Divide by -3:
$a = -\frac{11}{3}$

Answer:

$-\frac{11}{3}$