Solve: (2×4×6)/(7×8×9) Raised to Power 3x

Q: Why can't I distribute the exponent to each number separately?

The exponent 3x applies to the entire grouped expression $ (2\times4\times6) $, not individual factors. Distributing would change the mathematical meaning completely!

Q: What's the difference between this and expanding the powers?

This question asks for the equivalent expression using exponent rules, not for calculating actual values. Keep the structure $ \frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}} $ intact.

Q: Do I need to calculate 2×4×6 and 7×8×9 first?

No! The question wants you to show how the exponent rule applies. Leave the products as $ (2\times4\times6)^{3x} $ and $ (7\times8\times9)^{3x} $.

Q: How do I remember the power of a fraction rule?

Think: "Power goes to both floors!" The exponent must be applied to both the numerator (top) and denominator (bottom) of the fraction.

Exponent Rules with Fractional Expressions

Insert the corresponding expression:

$\left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}=$

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Understand the problem

Insert the corresponding expression:

$\left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}=$

Step-by-step solution

Let's analyze the expression we are given:

$\left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}$

The expression is a power of a fraction. The rule for powers of a fraction is that each component of the fraction must be raised to the power separately. This can be expressed as:

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

Applying this rule to our expression, we have:

The numerator inside the power: $2 \times 4 \times 6$
The denominator inside the power: $7 \times 8 \times 9$

Therefore, raising each part to the power $3x$ gives us:

$\frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}$

Thus, the simplified expression for the given equation is:

$\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

The solution to the question is: $\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

Final Answer

$\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

Key Points to Remember

Essential concepts to master this topic

Rule: Apply exponent to both numerator and denominator separately
Technique: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ transforms the entire expression
Check: Verify parentheses group all factors before applying exponent ✓

Common Mistakes

Avoid these frequent errors

Applying the exponent to individual factors instead of grouped terms
Don't apply 3x to each factor like $\frac{2^{3x} \times 4^{3x} \times 6^{3x}}{7^{3x} \times 8^{3x} \times 9^{3x}}$ = wrong structure! This misapplies the power rule and creates a different expression. Always keep the product grouped: $\frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}$ .

Practice Quiz

Test your knowledge with interactive questions

$112^0=\text{?}$

FAQ

Everything you need to know about this question

Why can't I distribute the exponent to each number separately?

The exponent 3x applies to the entire grouped expression $(2\times4\times6)$ , not individual factors. Distributing would change the mathematical meaning completely!

What's the difference between this and expanding the powers?

This question asks for the equivalent expression using exponent rules, not for calculating actual values. Keep the structure $\frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}$ intact.

Do I need to calculate 2×4×6 and 7×8×9 first?

No! The question wants you to show how the exponent rule applies. Leave the products as $(2\times4\times6)^{3x}$ and $(7\times8\times9)^{3x}$ .

How do I remember the power of a fraction rule?

Think: "Power goes to both floors!" The exponent must be applied to both the numerator (top) and denominator (bottom) of the fraction.

What if the exponent was just a number instead of 3x?

The same rule applies! Whether the exponent is 2, 3x, or n, you always apply it to both the numerator and denominator separately using $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ .

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