Complete the Expression: (7×5×2)^y Mathematical Power Problem

Power of Products with Multiple Factors

Insert the corresponding expression:

(7×5×2)y= \left(7\times5\times2\right)^y=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Simplify the following problem
00:04 In order to open parentheses with a multiplication operation and an outside exponent
00:08 We raise each factor to the power
00:15 We will apply this formula to our exercise
00:25 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Insert the corresponding expression:

(7×5×2)y= \left(7\times5\times2\right)^y=

2

Step-by-step solution

To solve the problem of expressing (7×5×2)y(7 \times 5 \times 2)^y in another form, we'll use the power of a product rule step by step:

Step 1: Identify and Understand the Given Expression
We are given (7×5×2)y(7 \times 5 \times 2)^y. This indicates that the entire product inside the parentheses is raised to the power yy.

Step 2: Apply the Power of a Product Rule
The power of a product rule states that (a×b×c)n=an×bn×cn\left(a \times b \times c\right)^n = a^n \times b^n \times c^n. According to this law, we can distribute the exponent yy to each factor within the parentheses.

Step 3: Rewrite the Expression
Applying this rule, we rewrite the expression as:

  • Raise the first factor, 7, to the power of yy: 7y7^y.
  • Raise the second factor, 5, to the power of yy: 5y5^y.
  • Raise the third factor, 2, to the power of yy: 2y2^y.

Putting it all together, the expression becomes 7y×5y×2y7^y \times 5^y \times 2^y.

Conclusion
Therefore, the expression (7×5×2)y(7 \times 5 \times 2)^y is correctly rewritten as 7y×5y×2y7^y \times 5^y \times 2^y.

Hence, the corresponding expression is 7y×5y×2y7^y \times 5^y \times 2^y.

3

Final Answer

5y×7y×5y 5^y\times7^y\times5^y

Key Points to Remember

Essential concepts to master this topic
  • Rule: Distribute exponent to each factor inside parentheses
  • Technique: (7×5×2)y=7y×5y×2y (7 \times 5 \times 2)^y = 7^y \times 5^y \times 2^y
  • Check: Each base appears once with same exponent y ✓

Common Mistakes

Avoid these frequent errors
  • Distributing exponent to only some factors
    Don't write (7×5×2)y=7y×5×2 (7 \times 5 \times 2)^y = 7^y \times 5 \times 2 = incomplete distribution! This ignores the power rule and gives incorrect results. Always apply the exponent to every single factor inside the parentheses.

Practice Quiz

Test your knowledge with interactive questions

\( (4^2)^3+(g^3)^4= \)

FAQ

Everything you need to know about this question

Why does the exponent go to each number separately?

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The power of a product rule says when you raise a product to a power, each factor gets raised to that power. Think of it like distributing: (abc)n=an×bn×cn (abc)^n = a^n \times b^n \times c^n .

What if one of the numbers was already raised to a power?

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You'd use the power of a power rule! For example, (72×5)y=72y×5y (7^2 \times 5)^y = 7^{2y} \times 5^y . Multiply the exponents for that factor.

Can I simplify the numbers inside first?

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Yes! You could calculate 7×5×2=70 7 \times 5 \times 2 = 70 first, giving you 70y 70^y . Both forms are correct, but the expanded form shows the individual factors clearly.

How do I remember which form is correct?

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Look for the pattern: same exponent on every factor. In 7y×5y×2y 7^y \times 5^y \times 2^y , each base has exponent y, which matches our original expression.

What if the exponent was a number instead of y?

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The rule works exactly the same! For instance, (7×5×2)3=73×53×23 (7 \times 5 \times 2)^3 = 7^3 \times 5^3 \times 2^3 . Just substitute the actual number for the variable.

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