Simplify the Expression: y³×x⁴(xyz)²×(yˣ)ʸ Step by Step

Exponent Laws with Mixed Algebraic Bases

Solve the following problem :

y3×x4(xyz)2×(yx)y= y^3\times x^4(xyz)^2\times(y^x)^y=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:14 Let's simplify this problem together.
00:20 If terms are multiplied inside a bracket with a power outside, raise each term to that power.
00:35 Now, we use the commutative law to rearrange the terms in this problem.
00:52 Remember, when you have a power to another power, multiply the powers.
00:57 And when multiplying powers with the same base, add the exponents.
01:02 Let's use this rule: add up the powers in our exercise.
01:06 Great job! We've found the solution.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following problem :

y3×x4(xyz)2×(yx)y= y^3\times x^4(xyz)^2\times(y^x)^y=

2

Step-by-step solution

First, let’s carefully expand the parentheses by applying two exponent laws.

The first law is the product of powers rule, which applies when an exponent is raised over a product of terms:

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

This law states that when an exponent is applied to parentheses containing a product of terms, the exponent is distributed to each factor inside the parentheses.

The second law we'll use is the power of a power law:

(am)n=amn (a^m)^n=a^{m\cdot n}

This law states that when an exponent is applied to a term that is already raised to a power (whether written with parentheses for clarity or not), the exponents are multiplied together.

Now, let’s return to the problem and first simplify the two terms with parentheses separately before handling the overall multiplication.

  1. The first from left to right is:

(xyz)2=x2y2z2 (xyz)^2=x^2\cdot y^2\cdot z^2

When we expanded the parentheses using the first law, we applied the exponent to each factor in the product inside the parentheses.

  1. The second from left to right is:

(yx)y=yxy (y^x)^y=y^{x y}

When we expanded the parentheses using the second law, we obtained the following:

y3x4(xyz)2(yx)y=y3x4x2y2z2yxy y^3\cdot x^4\cdot(xyz)^2\cdot (y^x)^y=y^3\cdot x^4\cdot x^2\cdot y^2\cdot z^2\cdot y^{xy}

When we used 1 and 2 mentioned above.

From here on we will no longer indicate the multiplication sign, but use the conventional writing form where placing terms next to each other means multiplication.

Now let's arrange the expression by bases using the multiplication commutative law:

y3x4x2y2z2yxy=y3y2yxyx4x2z2 y^3x^4x^2y^2z^2y^{xy}=y^3y^2y^{xy}x^4x^2z^2

Let's continue and simplify the expression by using the law of exponents for multiplication between terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n}

Note that this law is valid for any number of terms in multiplication and not just for two, for example for multiplication of three terms with identical bases we get:

amanak=am+nak=am+n+k a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}

When we used the above exponent law twice, we can also perform the same calculation for four terms in multiplication five and so on..

Let's apply this law to the problem:

y3y2yxyx4x2z2=y3+2+xyx4+2z2=y5+xyx6z2 y^3y^2y^{xy}x^4x^2z^2=y^{3+2+xy}x^{4+2}z^2=y^{5+xy}x^6z^2

When we used the above exponent law for multiplication between terms with identical bases only for terms with the same base.

We got the most simplified expression, so we're done.

Therefore the correct answer is C (when taking into account the commutative law of multiplication).

Important note:

It's worth understanding the reason for the power of a power law mentioned above (the second law), this law comes directly from the definition of exponents:

(am)n=amamam=am+m+m++m=amn (a^m)^n=a^m\cdot a^m\cdot\ldots\cdot a^m=a^{m+m+m+\cdots+m}=a^{m\cdot n}

When in the first stage we applied the definition of exponents to the term in parentheses and multiplied it by itself n times, then we applied the above law of exponents for multiplication between terms with identical bases and interpreted the multiplication between terms as a sum in the exponent,

Then we used the simple multiplication definition stating that if we connect a number to itself n times we can simply write it as multiplication, meaning:

m+m++m=mn m+m+\cdots+m=m\cdot n

And therefore we get that:

(am)n=amn (a^m)^n=a^{m\cdot n}

3

Final Answer

x6y5+xyz2 x^6y^{5+xy}z^2

Key Points to Remember

Essential concepts to master this topic
  • Product Rule: (abc)n=anbncn (abc)^n = a^n \cdot b^n \cdot c^n distributes exponents
  • Power Rule: (yx)y=yxy (y^x)^y = y^{xy} multiplies exponents together
  • Check: Group like bases and add exponents: y3y2yxy=y5+xy y^3 \cdot y^2 \cdot y^{xy} = y^{5+xy}

Common Mistakes

Avoid these frequent errors
  • Adding exponents when bases are different
    Don't combine x4 x^4 and y3 y^3 into (xy)7 (xy)^7 = wrong expression! Different bases stay separate in multiplication. Always group identical bases first, then add their exponents: x4x2=x6 x^4 \cdot x^2 = x^6 and y3y2yxy=y5+xy y^3 \cdot y^2 \cdot y^{xy} = y^{5+xy} .

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

Why does (yx)y (y^x)^y become yxy y^{xy} and not yx+y y^{x+y} ?

+

The power of a power rule says to multiply exponents, not add them! When you have (am)n (a^m)^n , you get amn a^{m \cdot n} . So (yx)y=yxy=yxy (y^x)^y = y^{x \cdot y} = y^{xy} .

How do I handle (xyz)2 (xyz)^2 when there are three variables?

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Apply the exponent to each factor inside the parentheses! (xyz)2=x2y2z2 (xyz)^2 = x^2 \cdot y^2 \cdot z^2 . The product rule works for any number of variables.

Can I simplify y5+xy y^{5+xy} any further?

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No - this is the most simplified form! Since the exponent 5+xy 5+xy contains both numbers and variables, you cannot combine or reduce it further.

What's the difference between x4x2 x^4 \cdot x^2 and x4+x2 x^4 + x^2 ?

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Huge difference! x4x2=x4+2=x6 x^4 \cdot x^2 = x^{4+2} = x^6 (multiplication = add exponents), but x4+x2 x^4 + x^2 cannot be simplified further because it's addition, not multiplication.

Why do we rearrange the terms by grouping the same bases together?

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Grouping makes it easier to apply exponent rules! The commutative property lets us rearrange: y3x4y2=y3y2x4 y^3x^4y^2 = y^3 \cdot y^2 \cdot x^4 . Now we can clearly see which exponents to add together.

How can I check if my final answer is correct?

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Verify by working backwards! Start with your answer x6y5+xyz2 x^6y^{5+xy}z^2 and see if you can reconstruct the original expression by reversing the exponent laws. The bases and total powers should match!

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