Simplify the Expression: x³ × 7x × 2x⁻³ Using Laws of Exponents

Exponent Laws with Negative Powers

x3×7x×2x3= x^3\times7x\times2x^{-3}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Simplify the following expression
00:03 When multiplying powers with equal bases
00:06 The power of the result equals the sum of powers
00:11 We'll apply this formula to our exercise and add together the powers
00:15 Let's solve the multiplication of numbers
00:31 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

x3×7x×2x3= x^3\times7x\times2x^{-3}=

2

Step-by-step solution

Here we have multiplication between all the elements in the expression, so we will first use the commutative law in multiplication approach the numbers separately, for clarity we will approach it in stages:

x37x2x3=72x3xx3=14x3xx3 x^3\cdot7x\cdot2x^{-3}=7\cdot2\cdot x^3\cdot x\cdot x^{-3}=14\cdot x^3\cdot x\cdot x^{-3} Note that it is possible (and even preferable) to skip the middle stage, meaning:

To write directly:x37x2x3=14x3xx3 x^3\cdot7x\cdot2x^{-3}=14\cdot x^3\cdot x\cdot x^{-3}

We will continue and use the associative law for multiplication between elements with the same bases:

cmcn=cm+n c^m\cdot c^n=c^{m+n} Note that this law is also valid for several elements in multiplication and not just for two, for example for a multiplication of three elements with the same base we will get:

cmcnck=cm+nck=cm+n+k c^m\cdot c^n\cdot c^k=c^{m+n}\cdot c^k=c^{m+n+k} When can use the associative even for four, five, or more elements in a multiplication.

Let's go back to the problem, and apply the associative law:

14x3xx3=14x3+13=14x1=14x 14x^3xx^{-3}=14x^{3+1-3}=14x^1=14x And therefore the correct answer is c.

Important note:

Here it is necessary to emphasize that you always need to ask the question - what do the parentheses apply to?

For example, in the problem here the parentheses only apply to the bases of the-

x x and not to the exponents, in a clearer way, also in the following expression:

5c7 5c^7 The parentheses apply only to c c and not to the exponent 5, as opposed to that when writing:

(5c)7 (5c)^7 The parentheses apply to each of the multiplication elements within the parentheses, meaning:

(5c)7=57c7 (5c)^7=5^7c^7 This is actually the application of the associative law:

(wr)n=wnrn (w\cdot r)^n=w^n\cdot r^n resulting both from the meaning of the parentheses and from the definition of parentheses.

3

Final Answer

14x 14x

Key Points to Remember

Essential concepts to master this topic
  • Product Rule: When multiplying same bases, add the exponents together
  • Technique: Rewrite x3xx3 x^3 \cdot x \cdot x^{-3} as x3+1+(3)=x1 x^{3+1+(-3)} = x^1
  • Check: Verify by substituting a test value like x=2: 237(2)2(23)=14 2^3 \cdot 7(2) \cdot 2(2^{-3}) = 14

Common Mistakes

Avoid these frequent errors
  • Adding exponents incorrectly with negative signs
    Don't compute 3 + 1 + (-3) as 3 + 1 - 3 = 1 without careful attention to signs = calculation errors! Students often drop negative signs or add instead of subtract. Always write out each step clearly: 3 + 1 + (-3) = 4 + (-3) = 1.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

Why do we multiply the coefficients 7 and 2 together?

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The commutative property lets us rearrange multiplication in any order! So x3×7x×2x3 x^3 \times 7x \times 2x^{-3} becomes 7×2×x3×x×x3=14x1 7 \times 2 \times x^3 \times x \times x^{-3} = 14x^1 .

What does x⁻³ actually mean?

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A negative exponent means reciprocal! So x3=1x3 x^{-3} = \frac{1}{x^3} . When multiplying, it's easier to just add exponents: x3×x3=x3+(3)=x0=1 x^3 \times x^{-3} = x^{3+(-3)} = x^0 = 1 .

How do I handle the x with no visible exponent?

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Remember that x=x1 x = x^1 ! The exponent 1 is usually invisible, but it's always there. So 7x=7x1 7x = 7x^1 when applying exponent rules.

Can I work with the variables and numbers separately?

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Absolutely! Group like terms: coefficients (7 × 2 = 14) and variables with same base (x3×x1×x3 x^3 \times x^1 \times x^{-3} ). This makes the problem much cleaner to solve.

What if I get x⁰ in my calculation?

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Great question! Remember that any base to the power of 0 equals 1 (except 0⁰). So x0=1 x^0 = 1 , which often simplifies your final answer significantly.

How can I check if 14x is really correct?

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Substitute any value for x (except 0). Try x = 1: 13×7(1)×2(13)=1×7×2×1=14 1^3 \times 7(1) \times 2(1^{-3}) = 1 \times 7 \times 2 \times 1 = 14 . And 14x=14(1)=14 14x = 14(1) = 14

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