Powers and Exponents Rules – Algebra Rules to Simplify Expressions with Powers and Exponents

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[argonwheatbelly]

48x⁹y⁷ in about 20 seconds staring at the screen. Just woke up. Nice one. I reviewed this kind of problem with some students the other day. It's the negative exponents that get them. I like the y⁰ red herring.

[franklinchinquee]

Wow! I've been out of high school for 55 years, never used this knowledge and it's all coming back now ... many thanks!?

[audreymcdonald]

Thanks for your sessions. I am a senior and i am really learnin

[russelllomando]

wow 48X^9Y^7 and now i'm done for the day my brain hurts thnaks for the fun

[dazartingstall]

Starting with the denominator:
(2²xy)⁻¹
= 2⁻²x⁻¹y⁻¹
= 0.25x⁻¹y⁻¹
The numerator:
y⁰ = 1 so we can ditch that at a glance. Which leaves us with:
(2x²y³)² × 3x⁴
= 4x⁴y⁶ × 3x⁴
= 12x⁸y⁶
So our fraction is:
12x⁸y⁶ ⁄ 0.25x⁻¹y⁻¹
= 48x⁹y⁷

[gavindeane]

The denominator is all to the power -1 so really we're just multiplying the numerator by 2²xy

The constant part is 2² × 3 × 2² = 48

Now emembering that
(x^a)^b = x^(ab)
and
(x^a)(x^b) = x^(a+b)


We have (x²)²x⁴x = x⁹

And we have (y³)²y⁰y = y⁷

So 48x⁹y⁷

[panlomito]

First y^0 = 1 so we can forget y^0...
Then (2x²y³)² = 4x^4 . y^6
And so 4x^4 . y^6 . 3x^4 . 4xy = 48 . x^9 . y^7

[dazartingstall]

Or we could go like this, just for the fun of it:
Simplify the numerator as before to 12x⁸y⁶
Then the denominator goes:
(2²xy)⁻¹ = 1/4xy
The problem, expressed as an in-line division becomes:
12x⁸y⁶ ÷ (1/4xy)
= 12x⁸y⁶ × 4xy
= 48x⁹y⁷

[Stylux-zp]

power rules :: (a^x) / (a^y) = (a^x-y) | (xyz)^2 = x^2y^2z^2 | a^m * a^n = a^m+n
((2x^2)(y^3))^2 (3x^4)(y^0) / ((2^2)xy)^-1
((2^2)(x^4)(y^6)) (3x^4)(y^0) / ((2^-1)(x^-1)(y^-1))
(3x^4)(y^0) = (2^4)(x^5)(y^7) (3x^4)(y^0)
(16x^5)(y^7) (3x^4)(y^0) = (48x^9)(y^7)

[chrisdissanayake]

(2X^2 Y^3)^2 3X^4 Y^0
———————————-
(2^2 XY)^-1

(2X^2 Y^3)(2X^2 Y^3)
= 4X^4 Y^6

3X^4 Y^0=3X^4*1
=3X^4

(2^2XY)^-1=
1/(2^2XY)= 1/4XY

4X^4 Y^6 * 3X^4
—————————-
1/ 4XY

4X^4 Y^6*3X^4*4XY

Answer: 48X^9 Y^7

[Y^0 = 1]

[venkataramanas.b.]

Almost most of the people got the answer within half a minute . It takes you almost 14 minutes.
😂😂😂😂😂

[tomtke]

First the problem
Numerator÷Denominator

Num=((2x^2•y^3)^2)(3x^4•y^0)
Denom=(2^2xy)^(-1)

Refresh:
A different example
(2^2)^3
=2^6?
(2^2)^3 = 4^3 = 64
2^6 = 64

Num=((2x^2•y^3)^2)(3x^4•y^0)
=(4x^4•y^6)(3x^4•1)
=12(x^8)(y^6)

Denom=(2^2xy)^(-1)
= 1/(4xy)

The approach to dividing by a fraction is to invert and multiply.

Num/Denom
= 12(x^8)(y^6)/(1/(4xy))
= 12(x^8)(y^6)•(4xy)
= 48(x^9)(y^7)

I'm not clear of how to verify simply.. so I'll guess at
x=3
y=5

Num=((2x^2•y^3)^2)(3x^4•y^0)
=((2•3^2•5^3)^2)(3•3^4•5^0)
=((18•125)^2)(3•81•1)
=((2, 250)^2)(243)
=(5, 062, 500)(243)
=1, 230, 187, 500✔️

Denom=(2^2xy)^(-1)
=(2^2•3•5)^(-1)
=(60)^(-1)
=1/60

Num/Denom
=1, 230, 187, 500 × 60
=73, 811, 250, 000❤

vs
= 48(x^9)(y^7)
= 48(3^9)(5^7)
= 48(19, 683)(78, 125)
= 48(1, 537, 734, 375)
= 73, 811, 250, 000❤

Final Answer
= 48(x^9)(y^7)