Question

Check exponents problem. WILL MEDAL!!!

I just need someone who will check some exponents problem. I solved the problems, just need someone to double check.

Answer 1

\[x^5 \times x^3 = x^8\]

Answer 2

@agent0smith @mathmale

Answer 3

Correct

Answer 4

\[\frac{6^5}{6^3} = 6^2 = 36\]

Answer 5

Correct

Answer 6

\[\frac{x^4}{x^7} = x^{-3} = \frac{1}{x^3}\]

Answer 7

yes, multiplying with exponents all you do is add the exponents then dividing with variables would be subtracting

Answer 8

Correct

Answer 9

\[(y^4)^3 = y^{12}\]

Answer 10

Correct

Answer 11

\[(x^2y)^4 = x^8y^4\]

Answer 12

Correct

Answer 13

\[\frac{8x^5}{4x^2} = 2x^3\]

Answer 14

Correct

Answer 15

\[\frac{x^5y^6}{xy^2} = x^4y^4\]

Answer 16

Correct

Answer 17

\[(\frac{5x^3y}{20xy^3})^4 = \frac{5^4x^{12}y^4}{20^4x^4y^{20}} = \frac{x^8}{4y^{16}}\]

Answer 18

\[7^{-2} = \frac{1}{49}\]

Answer 19

\[\frac{1}{x^{-5}} = x^5\]

Answer 20

\(x^{-6} = \frac{1}{x^6}\)

Answer 21

The last three are correct, but the complicated one is not.

Answer 22

How do we do that then?

Answer 23

Look at the exponent of the y in the denominator

Answer 24

Oh!

Answer 25

It was supposed to be \(y^{12}\) right?

Answer 26

Yes

Answer 27

Wait. I just looked at the original problem. It was supposed to be y^5. I am sorry.

Answer 28

:'(

Answer 29

I am sorry :'(

But would that answer be correct?

Answer 30

Meaning my original answer.

Answer 31

@skullpatrol

Answer 32

You should have 4^4 in the denominator

Answer 33

We would need to carry the exponent of 4?

Answer 34

Yes

Answer 35

Okay. So my answer would be \(\large \frac{x^8}{256y^{16}}\)

Answer 36

Correct the question please.

Answer 37

What do you mean?

Answer 38

y^5

Answer 39

\[\large (\frac{ 5x^3y }{ 20xy^5})^4 = \frac{ x^8 }{ 256y^{16} }\]

Answer 40

Correct :D

Answer 41

Okay moving on :)

Answer 42

\[x^9 \times x^{-7} = x^2\]

Answer 43

Correct

Answer 44

\[\frac{ x^{-1} }{ x^{-8} } = x^7\]

Answer 45

Correct

Answer 46

\[\frac{ x^{-4} }{ x^{-9} } = x^5\]

Answer 47

Correct

Answer 48

\[(2x^3y^{-3})^{-2} = \frac{y^6}{4x^6}\]

Answer 49

Correct

Answer 50

\[(4x^4y^{-4})^3 = \frac{164x^{12}}{y^{12}}\]

Answer 51

sorry i meant 64

Answer 52

\[\frac{ 64x^{12} }{ y^{12} }\]

Answer 53

Correct

Answer 54

\[(5x^2y)(2x^4y^{-3}) = \frac{10x^6}{y^3}\]

Answer 55

y^3 in the denominator?

Answer 56

Yes. Why?

Answer 57

What about the y in the numerator

Answer 58

What do you mean?

Answer 59

5x^2y

Answer 60

It's supposed to be a multiplication problem. I don't see why \(y\) needs to be in the numerator.

Answer 61

$$\Huge 5x^2y$$

Answer 62

I still don't understand. Please make it clear.

Answer 63

I just checked it out with Wolfram Alpha. They say that the denominator needs to have \(y^2\)?

Answer 64

$$\Huge y^{-3}=\dfrac{1}{y^3}$$

Answer 65

Oh!
\(\large y^1 \times y^{-3} = y^{1 + (-3)} = y^{-2} = \frac{1}{y^2}\)

Answer 66

Yes

Answer 67

OK got it! Thank you so much for your time and help! :)

Answer 68

Thanks for trying to learn :-)