Simplify: (1/x^7)(y^7) × Fourth Root of x^8

Question

1x7y7x84=? \frac{1}{x^7}\cdot y^7\cdot\sqrt[4]{x^8}=\text{?}

Video Solution

Solution Steps

00:00 Simply
00:03 To eliminate a negative exponent
00:06 Flip numerator and denominator and the exponent will become positive
00:09 We will use this formula in our exercise
00:23 Nth root of a number to the power of M
00:27 The result will be equal to the difference of exponents
00:30 We will use this formula in our exercise and subtract between the exponents
00:39 When multiplying powers with equal bases
00:43 The exponent of the result equals the sum of the exponents
00:50 We will use this formula in our exercise and sum the exponents
00:54 And this is the solution to the question

Step-by-Step Solution

First, let's deal with the root in the problem, we'll use the root and exponent law for this:

amn=(an)m=amn \sqrt[n]{a^m}=(\sqrt[n]{a})^m=a^{\frac{m}{n}}

Let's apply this exponent law to the problem:

1x7y7x84=1x7y7x84=1x7y7x2 \frac{1}{x^7}\cdot y^7\cdot\sqrt[4]{x^8}=\frac{1}{x^7}\cdot y^7\cdot x^{\frac{8}{4}}=\frac{1}{x^7}\cdot y^7\cdot x^2

When in the first stage we applied the above law for the third term in the product, we did this carefully while paying attention to what goes in the numerator of the fraction in the exponent? And what goes in the denominator of the fraction in the exponent? In the following stages, we simplified the expression we got,

Next, we'll recall the exponent law for negative exponents but in the opposite direction:

1an=an \frac{1}{a^n} =a^{-n}

And we'll apply this exponent law for the first term in the product in the expression we got in the last stage:

1x7y7x2=x7y7x2=y7x7x2 \frac{1}{x^7}\cdot y^7\cdot x^2=x^{-7}\cdot y^7\cdot x^2=y^7\cdot x^{-7}\cdot x^2

When in the first stage we applied the above exponent law for the first term in the product and in the next stage we arranged the expression we got using the commutative property of multiplication, so terms with identical bases would be adjacent to each other,

Next, we'll recall the exponent law for multiplying terms with identical bases:

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

We'll apply this exponent law to the expression we got in the last stage:

y7x7x2=y7x7+2=y7x5 y^7\cdot x^{-7}\cdot x^2=y^7x^{-7+2}=y^7x^{-5}

When in the first stage we applied the above exponent law for the terms with identical bases, and then simplified the expression we got, additionally in the final stages we removed the · sign and switched to the conventional notation where placing terms next to each other means multiplication.

Let's summarize the solution steps so far, we got that:

1x7y7x84=1x7y7x84=x7y7x2=y7x5 \frac{1}{x^7}\cdot y^7\cdot\sqrt[4]{x^8}=\frac{1}{x^7}\cdot y^7\cdot x^{\frac{8}{4}}=x^{-7}y^7x^2=y^7x^{-5}

Therefore, the correct answer is answer D.

Answer

y7x5 y^7x^{-5}