Simplify (x²×y³×z⁴)²: Advanced Exponent Expression

Power Rules with Multiple Variable Terms

Solve the following problem:

(x2×y3×z4)2= ? (x^2\times y^3\times z^4)^2=\text{ ?}

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1

Understand the problem

Solve the following problem:

(x2×y3×z4)2= ? (x^2\times y^3\times z^4)^2=\text{ ?}

2

Step-by-step solution

Let's solve this in two stages. In the first stage, we'll use the rule for power of a product in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

This rule states that, when raising a product in parentheses to a power, the power applies to each factor of the product when opening the parentheses.

Let's apply this rule to our problem:

(x2y3z4)2=(x2)2(y3)2(z4)2 (x^2\cdot y^3\cdot z^4)^2=(x^2)^2\cdot(y^3)^2\cdot(z^4)^2

In our case, when expanding the parentheses, we apply the power to each factor of the product separately. However, given that each of the factors are raised to a power, we do this carefully and use parentheses.

Next, we'll use the power rule for a power of a power:

(bm)n=bmn (b^m)^n=b^{m\cdot n}

Let's apply this law to the expression we have:

(x2)2(y3)2(z4)2=x22y32z42=x4y6z8 (x^2)^2\cdot(y^3)^2\cdot(z^4)^2=x^{2\cdot2}\cdot y^{3\cdot2}\cdot z^{4\cdot2}=x^4\cdot y^6\cdot z^8

In the second stage we perform the multiplication operation in the power exponents of the factors we obtained.

Therefore, the correct answer is answer D.

Note:

From the above formulation of the power law for parentheses, it might seem that it only refers to two factors in a product within parentheses, but in fact it is valid for a power of a product of multiple factors in parentheses, as demonstrated in this problem.

It would be a good exercise to prove that if the above law is valid for a power of a product of two factors in parentheses (as it is formulated above), then it is also valid for a power of a product of multiple factors in parentheses (for example - three factors, etc.).

3

Final Answer

x4y6z8 x^4y^6z^8

Key Points to Remember

Essential concepts to master this topic
  • Rule: When raising a product to a power, apply the power to each factor
  • Technique: Use (x2)2=x2×2=x4 (x^2)^2 = x^{2 \times 2} = x^4 for each variable
  • Check: Count total variables: x⁴y⁶z⁸ has 4+6+8=18 variable factors ✓

Common Mistakes

Avoid these frequent errors
  • Only applying the power to the first variable
    Don't just square x² to get x⁴ and forget the other variables = incomplete answer like x⁴! This ignores y³ and z⁴ completely. Always apply the outside power to every single factor inside the parentheses.

Practice Quiz

Test your knowledge with interactive questions

\( 112^0=\text{?} \)

FAQ

Everything you need to know about this question

Why do I multiply the exponents instead of adding them?

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When you have a power raised to another power, you multiply the exponents. This is different from adding exponents when multiplying same bases. Remember: (x2)3=x2×3=x6 (x^2)^3 = x^{2×3} = x^6

What's the difference between this and (xyz)²?

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Great question! (xyz)2=x2y2z2 (xyz)^2 = x^2y^2z^2 because each variable starts with power 1. But (x2y3z4)2=x4y6z8 (x^2y^3z^4)^2 = x^4y^6z^8 because we multiply the existing exponents by 2.

Do I need to write the multiplication signs?

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No! In algebra, x4y6z8 x^4y^6z^8 and x4×y6×z8 x^4 \times y^6 \times z^8 mean exactly the same thing. It's cleaner to write without the multiplication signs.

How can I remember to apply the power to every factor?

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Think of distribution! Just like distributing multiplication over addition, you must distribute the outside power to every single factor inside the parentheses. Count them to make sure you don't miss any!

What if there are numbers mixed with variables?

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Same rules apply! For example: (2x3y)4=24×x12×y4=16x12y4 (2x^3y)^4 = 2^4 \times x^{12} \times y^4 = 16x^{12}y^4 . Apply the power to both numbers and variables.

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