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

Q: Why does the negative exponent affect both 3 and a?

The parentheses make (3a) a single unit! The negative exponent -2 applies to the entire product inside the parentheses, not just individual factors. Think of it as $ (\text{whole thing})^{-2} $.

Q: Can I just make the exponent positive and keep everything the same?

No! Changing $ (3a)^{-2} $ to $ (3a)^2 $ gives you $ 9a^2 $, which is completely different from $ \frac{1}{9a^2} $. You must use the reciprocal rule!

Q: What if I wrote it as $ 3^{-2}a^{-2} $ first?

That's actually another valid approach! $ (3a)^{-2} = 3^{-2} \cdot a^{-2} = \frac{1}{3^2} \cdot \frac{1}{a^2} = \frac{1}{9a^2} $. Both methods give the same answer!

Q: Why is $ a \neq 0 $ important here?

When $ a = 0 $, the expression $ \frac{1}{9a^2} $ would have zero in the denominator, making it undefined. We need $ a \neq 0 $ to keep the fraction meaningful.

Q: How do I remember which rule to apply first?

Always handle negative exponents first! Convert $ (\text{expression})^{-n} $ to $ \frac{1}{(\text{expression})^n} $, then use other exponent rules on the positive exponent.

Negative Exponents with Product Expressions

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

$a\ne0$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Simplify the following expression

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

00:05 equals 1 divided by the number (A) raised 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 obtain 1 divided by (3A) squared

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

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

00:23 Let's apply this to the question

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

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

00:41 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

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

$a\ne0$

Step-by-step solution

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.

Final Answer

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

Key Points to Remember

Essential concepts to master this topic

Negative Exponent Rule: $b^{-n} = \frac{1}{b^n}$ flips to reciprocal
Power Rule: $(3a)^2 = 3^2 \cdot a^2 = 9a^2$ applies to products
Check: Substitute back: $\frac{1}{9a^2} \cdot 9a^2 = 1$ when multiplied by original ✓

Common Mistakes

Avoid these frequent errors

Applying negative exponent only to first factor
Don't write $(3a)^{-2} = \frac{1}{3}a^2$ = wrong result! This ignores that the negative exponent applies to the entire product (3a). Always apply the negative exponent rule first: $(3a)^{-2} = \frac{1}{(3a)^2}$ , then expand the denominator.

Practice Quiz

Test your knowledge with interactive questions

$$ (2^3)^6 = $$

FAQ

Everything you need to know about this question

Why does the negative exponent affect both 3 and a?

The parentheses make (3a) a single unit! The negative exponent -2 applies to the entire product inside the parentheses, not just individual factors. Think of it as $(\text{whole thing})^{-2}$ .

Can I just make the exponent positive and keep everything the same?

No! Changing $(3a)^{-2}$ to $(3a)^2$ gives you $9a^2$ , which is completely different from $\frac{1}{9a^2}$ . You must use the reciprocal rule!

What if I wrote it as $3^{-2}a^{-2}$ first?

That's actually another valid approach! $(3a)^{-2} = 3^{-2} \cdot a^{-2} = \frac{1}{3^2} \cdot \frac{1}{a^2} = \frac{1}{9a^2}$ . Both methods give the same answer!

Why is $a \neq 0$ important here?

When $a = 0$ , the expression $\frac{1}{9a^2}$ would have zero in the denominator, making it undefined. We need $a \neq 0$ to keep the fraction meaningful.

How do I remember which rule to apply first?

Always handle negative exponents first! Convert $(\text{expression})^{-n}$ to $\frac{1}{(\text{expression})^n}$ , then use other exponent rules on the positive exponent.

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

Negative Exponents with Product Expressions

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why does the negative exponent affect both 3 and a?

Can I just make the exponent positive and keep everything the same?

What if I wrote it as $3^{-2}a^{-2}$ first?

Why is $a \neq 0$ important here?

How do I remember which rule to apply first?

🌟 Unlock Your Math Potential

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

Negative Exponents with Product Expressions

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why does the negative exponent affect both 3 and a?

Can I just make the exponent positive and keep everything the same?

What if I wrote it as 3−2a−2 3^{-2}a^{-2} 3−2a−2 first?

Why is a≠0 a \neq 0 a=0 important here?

How do I remember which rule to apply first?

🌟 Unlock Your Math Potential

More Questions

Applying Combined Exponents Rules

Solve ((7×3)²)⁶ + (3⁻¹)³×(2³)⁴: Complex Exponent Challenge

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Evaluate (0.25)^(-2): Solving Negative Exponent Problems

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Solve (3×2×4×6)^(-4): Negative Exponent Calculation

Calculate (2×8×7)²: Square of Triple Product Problem

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Simplify the Expression: 7^5 × 7^(-6) Using Laws of Exponents

Calculate 7^(-4): Evaluating Negative Integer Exponents

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What if I wrote it as $3^{-2}a^{-2}$ first?

Why is $a \neq 0$ important here?