Solve (3a)^-2: Working with Negative Exponents Step by Step

Question

$(3a)^{-2}=\text{?}$

$a\ne0$

00:00 Simplify the expression

00:02 According to the laws of exponents, any number (A) to the power of (-N)

00:05 equals 1 divided by the number (A) to the power of (N)

00:08 Let's apply this to the question, the formula works from number to fraction and vice versa

00:11 We get 1 divided by (3A) squared

00:15 According to the laws of exponents, the number (A*B) to the power of (N)

00:19 equals (A) to the power of (N) multiplied by (B) to the power of (N)

00:23 Let's apply this to the question

00:26 We get (3) squared multiplied by (A) squared

00:33 We'll solve 3 squared according to the laws of exponents

00:41 And this is the solution to the question

We begin by using the negative exponent rule:

$b^{-n}=\frac{1}{b^n}$ We apply it to the given expression and obtain the following:

$(3a)^{-2}=\frac{1}{(3a)^2}$ We then use the power rule for parentheses:

$(x\cdot y)^n=x^n\cdot y^n$ We apply it to the denominator of the expression and obtain the following:

$\frac{1}{(3a)^2}=\frac{1}{3^2a^2}=\frac{1}{9a^2}$ Let's summarize the solution to the problem:

$(3a)^{-2}=\frac{1}{(3a)^2} =\frac{1}{9a^2}$

Therefore, the correct answer is option A.

$\frac{1}{9a^2}$